SDOI2015 Round1 Solution[完结]

dzy posted @ 2015年4月13日 10:07 in 专题 , 5577 阅读

作为一只不在SD省的蒟蒻,很有幸膜拜到了贵省的题目。

我就稍微讲讲做法吧。(其实还有代码反正都是为了骗访问量>_<)

 

D1T1 排序[OK]  D1T2 寻宝游戏[OK]  D1T3 序列统计[OK]

D2T1 星际战争[OK]  D2T2 约数个数和[OK]  D2T3 道路修建[OK]

 

终于推完了。。不过因为还没有得到数据,并不敢保证代码一定能AC。。。

[UPD 4.15]BZOJ上有数据了。好像都A了。

 


【D1T1 排序】

我们考虑已经使用了$1\sim i-1$这些操作中的某些操作后,判断能否继续排序以及是否需要第$i$次操作。

我们把序列分成$2^{n-i}$块,这样每段的长度都是$2^i$。我们定义一段是合法的,当且仅当这段数是从小到大排列的而且是连续的。如果发现有超过2段是不合法的,那这个序列显然就没救了,因为一次操作只能改变两段的状态。

如果一段是不合法的,那把它从中间剖开,形成的两段应该都是合法的,只有一种交换方法,判断是否可行。

假设有2段不合法,都从中间剖开,就得到了4段合法的序列。有4种交换方法,枚举是否可行即可。

因为操作的顺序是没有影响的,所以用了几次操作,最后答案就加上操作数量的阶乘即可。

这样我们就得到了一个搜索的程序,感觉合法的状态并没有很多,好像可以跑出极限数据,然而我并不清楚复杂度。


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
int n, bad;
vector<int> a;
typedef long long ll;
ll fact[15], ans;
void flip(vector<int> &A, int x, int y, int len){
	for(int i = 0;i < len;i ++)
		swap(A[x + i], A[y + i]);
}
int check(vector<int> A, int x, int y){
	for(int i = x + 1;i <= y;i ++)
		if(A[i] != A[i - 1] + 1) return 0;
	return 1;
}
void dfs(vector<int> A, int i, int cnt){
	if(i == n){
		ans += fact[cnt];
		return;
	}
	int b[5], tot = 0;
	for(int j = 0;j < (1 << n);j += (1 << (i + 1))){
		if(!check(A, j, j + (1 << (i + 1)) - 1)){
			if(tot == 4) return;
			b[++ tot] = j, b[++ tot] = j + (1 << i);
		}
	}
	vector<int> B;
	if(!tot) dfs(A, i + 1, cnt);
	if(tot == 2){
		if(A[b[2]] + (1 << i) == A[b[1]]){
			B = A;
			flip(B, b[1], b[2], 1 << i);
			dfs(B, i + 1, cnt + 1);
		}
	}
	if(tot == 4){
		vector<int> B;
		if(A[b[1]] + (1 << i) == A[b[3]] && A[b[2]] + (1 << i) == A[b[4]]){
			B = A;
			flip(B, b[2], b[3], 1 << i);
			dfs(B, i + 1, cnt + 1);
		}
		if(A[b[1]] + (1 << i) == A[b[4]] && A[b[3]] + (1 << i) == A[b[2]]){
			B = A;
			flip(B, b[2], b[4], 1 << i);
			dfs(B, i + 1, cnt + 1);
		}
		if(A[b[3]] + (1 << i) == A[b[2]] && A[b[1]] + (1 << i) == A[b[4]]){
			B = A;
			flip(B, b[1], b[3], 1 << i);
			dfs(B, i + 1, cnt + 1);
		}
		if(A[b[4]] + (1 << i) == A[b[2]] && A[b[3]] + (1 << i) == A[b[1]]){
			B = A;
			flip(B, b[1], b[4], 1 << i);
			dfs(B, i + 1, cnt + 1);
		}
	}
}

int main(){
	scanf("%d", &n);
	fact[0] = 1;
	for(int i = 1;i <= n;i ++)
		fact[i] = fact[i - 1] * i;
	a.resize(1 << n);
	for(int i = 0;i < (1 << n);i ++)
		scanf("%d", &a[i]);
	dfs(a, 0, 0);
	printf("%lld\n", ans);
	return 0;
}


【D1T2 寻宝游戏】

不知道大家有没有见过这题啊。。。

把加入的点按dfs序排序,把相邻的两个点以及头尾两个点的距离加起来就是答案了。(证明可以自己脑补)

那么用set维护这些点的位置就好了。

时间复杂度$O((n+m)\log n)$


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;

#define N 100010

typedef long long ll;

int h[N], g[N], p[N * 2], n1[N * 2], w[N * 2], ee;
int st[N], top, d[N], anc[20][N], n, pos[N], ptot, seq[N], ok[N];
ll dis[N], ans;

set <int> S;
typedef set <int> :: iterator SI;

void ae(int x, int y, int z){
	p[ee] = y; n1[ee] = h[x]; w[ee] = z; h[x] = ee ++;
}

void dfs(){
	memset(d, -1, sizeof(d));
	d[1] = 1;
	pos[1] = ++ ptot;
	seq[1] = 1;
	st[++ top] = 1;
	while(top){
		int u = st[top], flag = 0;
		for(int i = g[u];~i;i = n1[i]){
			if(!~d[p[i]]){
				pos[p[i]] = ++ ptot;
				seq[ptot] = p[i];
				st[++ top] = p[i];
				g[u] = n1[i];
				d[p[i]] = d[u] + 1;
				dis[p[i]] = dis[u] + w[i];
				anc[0][p[i]] = u;
				flag = 1;
				break;
			}
		}
		if(!flag) top --;
	}
}

int lca(int x, int y){
	if(d[x] < d[y]) swap(x, y);
	int k = d[x] - d[y], j = 0;
	while(k){
		if(k & 1) x = anc[j][x];
		k >>= 1;
		j ++;
	}
	if(x == y) return x;
	for(int i = 16;~i;i --) if(anc[i][x] != anc[i][y]) x = anc[i][x], y = anc[i][y];
	return anc[0][x];
}

ll dist(ll x, ll y){
	return dis[x] + dis[y] - 2 * dis[lca(x, y)];
}

int main(){
	int Q;
	scanf("%d%d", &n, &Q);
	memset(h, -1, sizeof(h));
	int x, y, z;
	for(int i = 1;i < n;i ++){
		scanf("%d%d%d", &x, &y, &z);
		ae(x, y, z);
		ae(y, x, z);
	}
	memcpy(g, h, sizeof(h));
	dfs();
	for(int j = 1;j <= 16;j ++)
		for(int i = 1;i <= n;i ++)
			anc[j][i] = anc[j - 1][anc[j - 1][i]];
	SI it, pre, suc;
	while(Q --){
		scanf("%d", &x);
		if(!ok[x]){
			ok[x] = 1;
			pre = suc = it = S.insert(pos[x]).first;
			if(S.size() == 2) ans = 2 * dist(seq[*S.begin()], seq[*S.rbegin()]);
			else if(S.size() > 2){
				if(pre == S.begin()) pre = S.end(), pre --;
				else pre --;
				suc ++;
				if(suc == S.end()) suc = S.begin();
				ans += dist(seq[*pre], x) + dist(seq[*suc], x) - dist(seq[*pre], seq[*suc]);
			}	
		}else{
			ok[x] = 0;
			pre = suc = it = S.find(pos[x]);
			if(S.size() <= 2){
				S.erase(pos[x]);
				ans = 0;	
			}else{
				if(pre == S.begin()) pre = S.end(), pre --;
				else pre --;
				suc ++;
				if(suc == S.end()) suc = S.begin();
				ans -= dist(seq[*pre], x) + dist(seq[*suc], x) - dist(seq[*pre], seq[*suc]);
				S.erase(pos[x]);
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}

	

【D1T3 序列统计】

求出$m$的原根$g$,然后对于每个$a_i$求出$g^{b_i}=a_i$。然后问题转化成了在序列$b$中,选出$n$个数(一个数可以取多次),且它们的和$s$满足$s \mod (m - 1) = x$。

用FFT+快速幂可以轻松做到$O(m\log m\log n)$的复杂度。


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define mo 1004535809
#define ww 3

#define N 300010

int g[20], ng[20], a[N], b[N], c[N], n, len, id[N], K, X, m, G, vis[N], vt, ni;

int pow(int a, int b, int c){
	int ret = 1;
	while(b){
		if(b & 1) ret = 1LL * ret * a % c;
		b >>= 1;
		a = 1LL * a * a % c;
	}
	return ret;
}

void FNT(int *A, int n, int flag){
	for(int i = n >> 1, j = 1;j < n;j ++){
		if(i < j) swap(A[i], A[j]);
		int k;
		for(k = n >> 1;i & k;i ^= k, k >>= 1);
		i ^= k;
	}
	int lg = 0;
	for(int m = 2;m <= n;m <<= 1){
		int w;
		++ lg;
		if(flag == 1) w = g[lg]; else w = ng[lg];
		for(int k = 0;k < n;k += m){
			int cur = 1;
			for(int j = k;j < k + (m >> 1);j ++, cur = 1LL * cur * w % mo){
				int u = A[j], t = 1LL * cur * A[j + (m >> 1)] % mo;
				A[j] = (u + t >= mo) ? (u + t - mo) : (u + t);
				A[j + (m >> 1)] = (u < t) ? (mo + u - t) : (u - t);
			}
		}
	}
}

void FFT_init(){
	for(int i = 1;i < 20;i ++){
		g[i] = pow(ww, (mo - 1) >> i, mo);
		ng[i] = pow(g[i], mo - 2, mo);
	}
}

int check(int k){
	int cur = 1;
	++ vt;
	for(int i = 1;i < n;i ++, cur = cur * k % n){
		if(vis[cur] == vt) return 0;
		vis[cur] = vt;
	}
	return 1;
}

int findroot(){
	for(int i = 2;i <= n;i ++)
		if(check(i)) return i;
}

int main(){
	scanf("%d%d%d%d", &K, &n, &X, &m);
	FFT_init();
	while((1 << len) <= 2 * n) len ++;
	ni = pow(1 << len, mo - 2, mo);
	G = findroot();
	int cur = 1;
	for(int i = 0;i < n - 1;i ++, cur = cur * G % n)
		id[cur] = i;
	int x;
	a[0] = 1;
	for(int i = 1;i <= m;i ++){
		scanf("%d", &x);
                if(x) b[id[x]] ++;
        }
	while(K){
		if(K & 1){
			FNT(a, 1 << len, 1);
			memcpy(c, b, sizeof(b));
			FNT(c, 1 << len, 1);
			for(int i = 0;i < (1 << len);i ++)
				a[i] = 1LL * a[i] * c[i] % mo;
			FNT(a, 1 << len, -1);
			for(int i = 0;i < (1 << len);i ++)
				a[i] = 1LL * a[i] * ni % mo;
			for(int i = n - 1;i < 2 * n - 1;i ++){
				a[i - n + 1] = (a[i - n + 1] + a[i]) % mo;
				a[i] = 0;
			}
		}
		K >>= 1;
		FNT(b, 1 << len, 1);
		for(int i = 0;i < (1 << len);i ++)
			b[i] = 1LL * b[i] * b[i] % mo;
		FNT(b, 1 << len, -1);
		for(int i = 0;i < (1 << len);i ++)
			b[i] = 1LL * b[i] * ni % mo;
		for(int i = n - 1;i < 2 * n - 1;i ++){
			b[i - n + 1] = (b[i - n + 1] + b[i]) % mo;
			b[i] = 0;
		}
	}
	printf("%d\n", a[id[X]]);
	return 0;
}

【D2T1 星际战争】

二分答案,然后弄两排点,能攻击的就连边,求最大流判断是否可行嘛。太裸了吧。


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int h[105], n1[10000], p[10000], ee, d[105], q[105], S, T, a[55], b[55], g[51][51], n, m;
double c[10000], mid;
const double eps = 1e-8;

void ae(int x, int y, double z){
	p[ee] = y; n1[ee] = h[x]; c[ee] = z; h[x] = ee ++;
	p[ee] = x; n1[ee] = h[y]; c[ee] = 0; h[y] = ee ++;
}

int bfs(){
	int s = 1, e = 1;
	memset(d, -1, sizeof(d));
	d[q[1] = S] = 0;
	while(s <= e){
		int u = q[s ++];
		for(int i = h[u];~i;i = n1[i])
			if(c[i] > 0 && !~d[p[i]])
				q[++ e] = p[i], d[p[i]] = d[u] + 1;
	}
	return d[T] != -1;
}

double dfs(int u, double lmt){
	if(u == T) return lmt;
	double ret = 0, del;
	for(int i = h[u];~i;i = n1[i]){
		if(c[i] > 0 && d[p[i]] == d[u] + 1){
			del = dfs(p[i], min(c[i], lmt - ret));
			c[i] -= del;
			c[i ^ 1] += del;
			ret += del;
			if(fabs(ret - lmt) < eps) return ret;
		}
	}
	if(fabs(ret) < eps) d[u] = -1;
	return ret;
}

double dinic(){
	double ret = 0;
	while(bfs()) ret += dfs(S, 1e10);
	return ret;
}

int check(){
	memset(h, -1, sizeof(h));
	ee = 0;
	S = 0;
	T = n + m + 1;
	int sum = 0;
	for(int i = 1;i <= m;i ++)
		ae(S, i, mid * b[i]);
	for(int i = 1;i <= n;i ++)
		ae(m + i, T, a[i]), sum += a[i];
	for(int i = 1;i <= m;i ++)
		for(int j = 1;j <= n;j ++)
			if(g[i][j])
				ae(i, m + j, 1e10);
	return dinic() >= sum;
}

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1;i <= n;i ++)
		scanf("%d", &a[i]);
	for(int i = 1;i <= m;i ++)
		scanf("%d", &b[i]);
	for(int i = 1;i <= m;i ++)
		for(int j = 1;j <= n;j ++)
			scanf("%d", &g[i][j]);
	double l = 0, r = 5e6;
	while(l + 1e-5 < r){
		mid = (l + r) / 2;
		if(check()) r = mid; else l = mid;
	}
	printf("%.4lf\n", mid);
	return 0;
}

【D2T2 约数个数和】

看到这个应该立马想到CF 235E。。。

[tex]\sum_{i=1}^{n} \sum_{j=1}^{m} d(ij)=\sum_{i=1}^{n} \sum_{j=1}^{m}\lfloor \frac{n}{i} \rfloor \lfloor \frac{m}{j} \rfloor [\gcd(i,j)==1][/tex]

[tex]=\sum_{i=1}^{n}\sum_{j=1}^{m}\lfloor \frac{n}{i} \rfloor \lfloor \frac{m}{j} \rfloor\sum_{k|\gcd(i,j)}\mu(k)[/tex]

[tex]=\sum_{i=1}^{n}\sum_{j=1}^{m}\lfloor \frac{n}{i} \rfloor \lfloor \frac{m}{j} \rfloor\sum_{k|i,k|j}\mu(k)[/tex]

[tex]=\sum_{k}\mu(k)\sum_{i=1}^{\lfloor \frac{n}{k} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{k} \rfloor}\lfloor \frac{n}{ik} \rfloor\lfloor \frac{m}{jk} \rfloor[/tex]

[tex]=\sum_{k}\mu(k)f(\lfloor \frac{n}{k} \rfloor)f(\lfloor \frac{m}{k} \rfloor)[/tex],其中[tex]f(n)=\sum_{i=1}^{n}\lfloor \frac{n}{i} \rfloor=\sum_{i=1}^{n}d(i)[/tex]

那么我们可以通过线性筛筛出约数个数$d$和$\mu$,询问的时候根号加速。这样整个复杂度就是$O(n+T\sqrt{n})$。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define N 50010

typedef long long ll;
int fl[N], pr[N], miu[N], f[N], g[N], cnt;

void sieve(int n){
	miu[1] = f[1] = 1;
	for(int i = 2;i <= n;i ++){
		if(!fl[i]) pr[++ cnt] = i, miu[i] = -1, f[i] = 2, g[i] = 1;
		for(int j = 1;j <= cnt && i * pr[j] <= n;j ++){
			fl[i * pr[j]] = 1;
			if(i % pr[j] == 0){
				miu[i * pr[j]] = 0;
				g[i * pr[j]] = g[i] + 1;
				f[i * pr[j]] = f[i] / (g[i] + 1) * (g[i] + 2);
				break;
			}
			miu[i * pr[j]] = -miu[i];
			f[i * pr[j]] = f[i] * 2;
			g[i * pr[j]] = 1;
		}
	}
	for(int i = 2;i <= n;i ++)
		f[i] += f[i - 1], miu[i] += miu[i - 1];
}

int main(){
	sieve(N - 10);
	int tc, n, m;
	scanf("%d", &tc);
	while(tc --){
		scanf("%d%d", &n, &m);
		if(n > m) swap(n, m);
		ll ans = 0;
		for(int i = 1, j;i <= n;i = j + 1){
			j = min(n / (n / i), m / (m / i));
			ans += 1LL * (miu[j] - miu[i - 1]) * f[n / i] * f[m / i];
		}
		printf("%lld\n", ans);
	}
	return 0;
}

【D2T3 道路修建】

(听说分治LCT可以写可惜我太弱还没去补过姿势)

[UPD 4.15 听说我这个做法太naive了。被吊打了。。。]

感觉很像BZOJ 1018 堵塞的交通啊。

我们考虑用线段树来维护嘛。

合并的时候,我们就是把两个高度为2的矩形内的东西合并起来。具体的说,我们只要考虑这个矩形四个角的联通情况。我们可以暴力枚举连通状态,然后用最小表示法表示,发现只有10个状态。

看看上面这图感受一下,标号相同表示他们在一个连通块内。

合并的时候,我们考虑加上两个子区间中间的边,即下图中红色方框的四条边。

那么我们可以枚举红边的子集,和左右两边的状态,进行暴力合并。每次复杂度10*10*2^4*合并的复杂度……(你TM在逗我笑?)

愉快的发现TLE了。

怎么办?我们可以预处理啊!我们枚举所有的转移状态$(x,y,z,state)$,表示左子树状态$x$和右子树状态$y$加上$state$边集可以合并起来得到状态$z$。然后把这些转移打在表里。发现只有600多种转移!这样复杂度是$O(600n\log n)$。感觉有点虚!

再仔细观察,有的状态是重复的!比如$(x,y)$加上任意3条边,都能转移到$z$。这样我们新建一种转移方式,原来3个转移可以缩减成1个!

经过优化,只剩300个不到个转移了。可以在3S内轻松出解。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 60010
typedef long long ll;
struct node{
	int w[2], cl, cr, f[11];
	void clear(){
		memset(f, 63, sizeof(f));
	}
}t[N << 2];

struct state{
	int x, y, z, s;
}e[300];
#define ls (x << 1)
#define rs (x << 1 | 1)

int W[2][N], C[N], n, Q, num = 266, wsum[18];

void upd(ll &x, ll y){
	x = min(x, y);
}

node merge(node l, node r){
	node x;
	x.w[0] = r.w[0];
	x.w[1] = r.w[1];
	x.cl = l.cl;
	x.cr = r.cr;
	for(int i = 1;i < 15;i ++){
		wsum[i] = 0;
		if(i & 1) wsum[i] += l.w[0];
		if(i & 2) wsum[i] += l.w[1];
		if(i & 4) wsum[i] += l.cr;
		if(i & 8) wsum[i] += r.cl;
	}
	wsum[0] = wsum[14] + l.w[0];
	wsum[16] = l.cr + min(l.w[0], l.w[1]);
	wsum[17] = r.cl + min(l.w[0], l.w[1]);
	x.clear();
	for(int i = 1;i <= num;i ++)
		x.f[e[i].z] = min(x.f[e[i].z], l.f[e[i].x] + r.f[e[i].y] + wsum[e[i].s]);
	return x;
}

void build(int l, int r, int x){
	if(l == r){
		t[x].clear();
		t[x].f[0] = C[l];
		t[x].f[9] = 0;
		t[x].w[0] = W[0][l];
		t[x].w[1] = W[1][l];
		t[x].cl = t[x].cr = C[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(l, mid, ls);
	build(mid + 1, r, rs);
	t[x] = merge(t[ls], t[rs]);
}

node ask(int ql, int qr, int l, int r, int x){
	if(ql <= l && r <= qr) return t[x];
	int mid = (l + r) >> 1;
	if(ql <= mid && mid >= qr) return ask(ql, qr, l, mid, ls);
	if(mid < qr && ql > mid) return ask(ql, qr, mid + 1, r, rs);
	return merge(ask(ql, qr, l, mid, ls), ask(ql, qr, mid + 1, r, rs));
}

void modifyc(int p, int v, int l, int r, int x){
	if(l == r){
		t[x].cl = t[x].cr = v;
		t[x].f[0] = v;
		return;
	}
	int mid = (l + r) >> 1;
	if(p <= mid) modifyc(p, v, l, mid, ls); else modifyc(p, v, mid + 1, r, rs);
	t[x] = merge(t[ls], t[rs]);
}

void modifyw(int p, int v, int vv, int l, int r, int x){
	if(l == r){
		t[x].w[v] = vv;
		return;
	}
	int mid = (l + r) >> 1;
	if(p <= mid) modifyw(p, v, vv, l, mid, ls); else modifyw(p, v, vv, mid + 1, r, rs);
	t[x] = merge(t[ls], t[rs]);
}

void init(){
	e[1] = (state){0, 0, 0, 0}; e[2] = (state){0, 0, 0, 16}; e[3] = (state){0, 0, 0, 17}; e[3] = (state){0, 0, 0, 1}; e[4] = (state){0, 0, 0, 2}; e[5] = (state){0, 0, 0, 3}; e[6] = (state){0, 1, 0, 0}; e[7] = (state){0, 1, 0, 17}; e[7] = (state){0, 1, 0, 3}; e[8] = (state){0, 2, 2, 0}; e[9] = (state){0, 2, 2, 16}; e[10] = (state){0, 2, 2, 17}; e[10] = (state){0, 2, 2, 1}; e[11] = (state){0, 2, 2, 2}; e[12] = (state){0, 2, 2, 3}; e[13] = (state){0, 3, 0, 0}; e[14] = (state){0, 3, 0, 17}; e[14] = (state){0, 3, 0, 3}; e[15] = (state){0, 4, 4, 0}; e[16] = (state){0, 4, 4, 16}; e[17] = (state){0, 4, 4, 17}; e[17] = (state){0, 4, 4, 1}; e[18] = (state){0, 4, 4, 2}; e[19] = (state){0, 4, 4, 3}; e[20] = (state){0, 5, 2, 0}; e[21] = (state){0, 5, 2, 17}; e[21] = (state){0, 5, 2, 3}; e[22] = (state){0, 6, 4, 0}; e[23] = (state){0, 6, 4, 17}; e[23] = (state){0, 6, 4, 3}; e[24] = (state){0, 7, 2, 0}; e[25] = (state){0, 7, 2, 17}; e[25] = (state){0, 7, 2, 3}; e[26] = (state){0, 8, 4, 0}; e[27] = (state){0, 8, 4, 17}; e[27] = (state){0, 8, 4, 3}; e[28] = (state){0, 9, 0, 0}; e[29] = (state){0, 9, 0, 17}; e[29] = (state){0, 9, 4, 1}; e[30] = (state){0, 9, 2, 2}; e[31] = (state){0, 9, 0, 3}; e[32] = (state){0, 9, 4, 5}; e[33] = (state){0, 9, 2, 6}; e[34] = (state){1, 0, 1, 0}; e[35] = (state){1, 0, 1, 16}; e[36] = (state){1, 0, 1, 17}; e[36] = (state){1, 0, 1, 1}; e[37] = (state){1, 0, 1, 2}; e[38] = (state){1, 0, 1, 3}; e[39] = (state){1, 1, 1, 0}; e[40] = (state){1, 1, 1, 17}; e[40] = (state){1, 1, 1, 3}; e[41] = (state){1, 2, 7, 0}; e[42] = (state){1, 2, 7, 16}; e[43] = (state){1, 2, 7, 17}; e[43] = (state){1, 2, 7, 1}; e[44] = (state){1, 2, 7, 2}; e[45] = (state){1, 2, 7, 3}; e[46] = (state){1, 3, 1, 0}; e[47] = (state){1, 3, 1, 17}; e[47] = (state){1, 3, 1, 3}; e[48] = (state){1, 4, 6, 0}; e[49] = (state){1, 4, 6, 16}; e[50] = (state){1, 4, 6, 17}; e[50] = (state){1, 4, 6, 1}; e[51] = (state){1, 4, 6, 2}; e[52] = (state){1, 4, 6, 3}; e[53] = (state){1, 5, 7, 0}; e[54] = (state){1, 5, 7, 17}; e[54] = (state){1, 5, 7, 3}; e[55] = (state){1, 6, 6, 0}; e[56] = (state){1, 6, 6, 17}; e[56] = (state){1, 6, 6, 3}; e[57] = (state){1, 7, 7, 0}; e[58] = (state){1, 7, 7, 17}; e[58] = (state){1, 7, 7, 3}; e[59] = (state){1, 8, 6, 0}; e[60] = (state){1, 8, 6, 17}; e[60] = (state){1, 8, 6, 3}; e[61] = (state){1, 9, 1, 0}; e[62] = (state){1, 9, 1, 17}; e[62] = (state){1, 9, 6, 1}; e[63] = (state){1, 9, 7, 2}; e[64] = (state){1, 9, 1, 3}; e[65] = (state){1, 9, 6, 5}; e[66] = (state){1, 9, 7, 6}; e[67] = (state){2, 0, 0, 0}; e[68] = (state){2, 0, 0, 16}; e[69] = (state){2, 0, 0, 3}; e[70] = (state){2, 1, 0, 0}; e[71] = (state){2, 2, 2, 0}; e[72] = (state){2, 2, 2, 16}; e[73] = (state){2, 2, 2, 3}; e[74] = (state){2, 3, 0, 0}; e[75] = (state){2, 4, 4, 0}; e[76] = (state){2, 4, 4, 16}; e[77] = (state){2, 4, 4, 3}; e[78] = (state){2, 5, 2, 0}; e[79] = (state){2, 6, 4, 0}; e[80] = (state){2, 7, 2, 0}; e[81] = (state){2, 8, 4, 0}; e[82] = (state){2, 9, 0, 0}; e[83] = (state){2, 9, 2, 3}; e[84] = (state){2, 9, 4, 5}; e[85] = (state){2, 9, 2, 6}; e[86] = (state){3, 0, 3, 0}; e[87] = (state){3, 0, 3, 16}; e[88] = (state){3, 0, 3, 17}; e[88] = (state){3, 0, 3, 1}; e[89] = (state){3, 0, 3, 2}; e[90] = (state){3, 0, 3, 3}; e[91] = (state){3, 1, 3, 0}; e[92] = (state){3, 1, 3, 17}; e[92] = (state){3, 1, 3, 3}; e[93] = (state){3, 2, 5, 0}; e[94] = (state){3, 2, 5, 16}; e[95] = (state){3, 2, 5, 17}; e[95] = (state){3, 2, 5, 1}; e[96] = (state){3, 2, 5, 2}; e[97] = (state){3, 2, 5, 3}; e[98] = (state){3, 3, 3, 0}; e[99] = (state){3, 3, 3, 17}; e[99] = (state){3, 3, 3, 3}; e[100] = (state){3, 4, 8, 0}; e[101] = (state){3, 4, 8, 16}; e[102] = (state){3, 4, 8, 17}; e[102] = (state){3, 4, 8, 1}; e[103] = (state){3, 4, 8, 2}; e[104] = (state){3, 4, 8, 3}; e[105] = (state){3, 5, 5, 0}; e[106] = (state){3, 5, 5, 17}; e[106] = (state){3, 5, 5, 3}; e[107] = (state){3, 6, 8, 0}; e[108] = (state){3, 6, 8, 17}; e[108] = (state){3, 6, 8, 3}; e[109] = (state){3, 7, 5, 0}; e[110] = (state){3, 7, 5, 17}; e[110] = (state){3, 7, 5, 3}; e[111] = (state){3, 8, 8, 0}; e[112] = (state){3, 8, 8, 17}; e[112] = (state){3, 8, 8, 3}; e[113] = (state){3, 9, 3, 0}; e[114] = (state){3, 9, 3, 17}; e[114] = (state){3, 9, 8, 1}; e[115] = (state){3, 9, 5, 2}; e[116] = (state){3, 9, 3, 3}; e[117] = (state){3, 9, 8, 5}; e[118] = (state){3, 9, 5, 6}; e[119] = (state){4, 0, 0, 0}; e[120] = (state){4, 0, 0, 16}; e[121] = (state){4, 0, 0, 3}; e[122] = (state){4, 1, 0, 0}; e[123] = (state){4, 2, 2, 0}; e[124] = (state){4, 2, 2, 16}; e[125] = (state){4, 2, 2, 3}; e[126] = (state){4, 3, 0, 0}; e[127] = (state){4, 4, 4, 0}; e[128] = (state){4, 4, 4, 16}; e[129] = (state){4, 4, 4, 3}; e[130] = (state){4, 5, 2, 0}; e[131] = (state){4, 6, 4, 0}; e[132] = (state){4, 7, 2, 0}; e[133] = (state){4, 8, 4, 0}; e[134] = (state){4, 9, 0, 0}; e[135] = (state){4, 9, 4, 3}; e[136] = (state){4, 9, 4, 5}; e[137] = (state){4, 9, 2, 6}; e[138] = (state){5, 0, 3, 0}; e[139] = (state){5, 0, 3, 16}; e[140] = (state){5, 0, 3, 3}; e[141] = (state){5, 1, 3, 0}; e[142] = (state){5, 2, 5, 0}; e[143] = (state){5, 2, 5, 16}; e[144] = (state){5, 2, 5, 3}; e[145] = (state){5, 3, 3, 0}; e[146] = (state){5, 4, 8, 0}; e[147] = (state){5, 4, 8, 16}; e[148] = (state){5, 4, 8, 3}; e[149] = (state){5, 5, 5, 0}; e[150] = (state){5, 6, 8, 0}; e[151] = (state){5, 7, 5, 0}; e[152] = (state){5, 8, 8, 0}; e[153] = (state){5, 9, 3, 0}; e[154] = (state){5, 9, 5, 3}; e[155] = (state){5, 9, 8, 5}; e[156] = (state){5, 9, 5, 6}; e[157] = (state){6, 0, 1, 0}; e[158] = (state){6, 0, 1, 16}; e[159] = (state){6, 0, 1, 3}; e[160] = (state){6, 1, 1, 0}; e[161] = (state){6, 2, 7, 0}; e[162] = (state){6, 2, 7, 16}; e[163] = (state){6, 2, 7, 3}; e[164] = (state){6, 3, 1, 0}; e[165] = (state){6, 4, 6, 0}; e[166] = (state){6, 4, 6, 16}; e[167] = (state){6, 4, 6, 3}; e[168] = (state){6, 5, 7, 0}; e[169] = (state){6, 6, 6, 0}; e[170] = (state){6, 7, 7, 0}; e[171] = (state){6, 8, 6, 0}; e[172] = (state){6, 9, 1, 0}; e[173] = (state){6, 9, 6, 3}; e[174] = (state){6, 9, 6, 5}; e[175] = (state){6, 9, 7, 6}; e[176] = (state){7, 0, 1, 0}; e[177] = (state){7, 0, 1, 16}; e[178] = (state){7, 0, 1, 3}; e[179] = (state){7, 1, 1, 0}; e[180] = (state){7, 2, 7, 0}; e[181] = (state){7, 2, 7, 16}; e[182] = (state){7, 2, 7, 3}; e[183] = (state){7, 3, 1, 0}; e[184] = (state){7, 4, 6, 0}; e[185] = (state){7, 4, 6, 16}; e[186] = (state){7, 4, 6, 3}; e[187] = (state){7, 5, 7, 0}; e[188] = (state){7, 6, 6, 0}; e[189] = (state){7, 7, 7, 0}; e[190] = (state){7, 8, 6, 0}; e[191] = (state){7, 9, 1, 0}; e[192] = (state){7, 9, 7, 3}; e[193] = (state){7, 9, 6, 5}; e[194] = (state){7, 9, 7, 6}; e[195] = (state){8, 0, 3, 0}; e[196] = (state){8, 0, 3, 16}; e[197] = (state){8, 0, 3, 3}; e[198] = (state){8, 1, 3, 0}; e[199] = (state){8, 2, 5, 0}; e[200] = (state){8, 2, 5, 16}; e[201] = (state){8, 2, 5, 3}; e[202] = (state){8, 3, 3, 0}; e[203] = (state){8, 4, 8, 0}; e[204] = (state){8, 4, 8, 16}; e[205] = (state){8, 4, 8, 3}; e[206] = (state){8, 5, 5, 0}; e[207] = (state){8, 6, 8, 0}; e[208] = (state){8, 7, 5, 0}; e[209] = (state){8, 8, 8, 0}; e[210] = (state){8, 9, 3, 0}; e[211] = (state){8, 9, 8, 3}; e[212] = (state){8, 9, 8, 5}; e[213] = (state){8, 9, 5, 6}; e[214] = (state){9, 0, 0, 0}; e[215] = (state){9, 0, 0, 16}; e[216] = (state){9, 0, 3, 1}; e[217] = (state){9, 0, 1, 2}; e[218] = (state){9, 0, 0, 3}; e[219] = (state){9, 0, 3, 9}; e[220] = (state){9, 0, 1, 10}; e[221] = (state){9, 1, 0, 0}; e[222] = (state){9, 1, 1, 3}; e[223] = (state){9, 1, 3, 9}; e[224] = (state){9, 1, 1, 10}; e[225] = (state){9, 2, 2, 0}; e[226] = (state){9, 2, 2, 16}; e[227] = (state){9, 2, 5, 1}; e[228] = (state){9, 2, 7, 2}; e[229] = (state){9, 2, 2, 3}; e[230] = (state){9, 2, 5, 9}; e[231] = (state){9, 2, 7, 10}; e[232] = (state){9, 3, 0, 0}; e[233] = (state){9, 3, 3, 3}; e[234] = (state){9, 3, 3, 9}; e[235] = (state){9, 3, 1, 10}; e[236] = (state){9, 4, 4, 0}; e[237] = (state){9, 4, 4, 16}; e[238] = (state){9, 4, 8, 1}; e[239] = (state){9, 4, 6, 2}; e[240] = (state){9, 4, 4, 3}; e[241] = (state){9, 4, 8, 9}; e[242] = (state){9, 4, 6, 10}; e[243] = (state){9, 5, 2, 0}; e[244] = (state){9, 5, 5, 3}; e[245] = (state){9, 5, 5, 9}; e[246] = (state){9, 5, 7, 10}; e[247] = (state){9, 6, 4, 0}; e[248] = (state){9, 6, 6, 3}; e[249] = (state){9, 6, 8, 9}; e[250] = (state){9, 6, 6, 10}; e[251] = (state){9, 7, 2, 0}; e[252] = (state){9, 7, 7, 3}; e[253] = (state){9, 7, 5, 9}; e[254] = (state){9, 7, 7, 10}; e[255] = (state){9, 8, 4, 0}; e[256] = (state){9, 8, 8, 3}; e[257] = (state){9, 8, 8, 9}; e[258] = (state){9, 8, 6, 10}; e[259] = (state){9, 9, 0, 0}; e[260] = (state){9, 9, 8, 1}; e[261] = (state){9, 9, 7, 2}; e[262] = (state){9, 9, 9, 3}; e[263] = (state){9, 9, 4, 5}; e[264] = (state){9, 9, 2, 6}; e[265] = (state){9, 9, 3, 9}; e[266] = (state){9, 9, 1, 10};
}

int main(){
	scanf("%d%d", &n, &Q);
	for(int j = 0;j < 2;j ++)
		for(int i = 1;i < n;i ++)
			scanf("%d", &W[j][i]);
	for(int i = 1;i <= n;i ++)
		scanf("%d", &C[i]);
	init();
	build(1, n, 1);
	int l, r, x1, y1, x2, y2, ww;
	char op[2];
	while(Q --){
		scanf("%s", op);
		if(op[0] == 'Q'){
			scanf("%d%d", &l, &r);
			printf("%d\n", ask(l, r, 1, n, 1).f[0]);
		}else{
			scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &ww);
			if(y1 == y2) modifyc(y1, ww, 1, n, 1);
			else modifyw(min(y1, y2), x1 - 1, ww, 1, n, 1);
		}
	}
	return 0;
}
  • 无匹配
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YZ23 说:
2015年6月11日 19:54

您好,请问D1T3要对卷积的系数取模,如何只用FFT而不用NTT?
谢谢~

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jackjohnny 说:
2021年9月14日 21:28

I admit, I have not been on this web page in a long time... however it was another joy to see It is such an important topic and ignored by so many, even professionals. I thank you to help making people more aware of possible issues. ไฮโล วิธีเล่น

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link 说:
2021年10月09日 16:47

This is a smart blog. I mean it. You have so much knowledge about this issue, and so much passion. You also know how to make people rally behind it, obviously from the responses. soap2day movies

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south university stu 说:
2022年8月26日 00:02

Student Portal Homepage. If you require assistance with an Existing Username or Password, please call the Technical Support Center: South University Online 2021. Welcome to Brightspace by D2L. Log in to view your courses, south university student login explore tools and features, and customize your e-Learning experience. Username Password.Student Portal Homepage. If you require assistance with an Existing Username or Password, please call the Technical Support Center: South University Online 2021. Welcome to Brightspace by D2L. Log in to view your courses.

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AP SSC General Scien 说:
2022年9月16日 21:49

All Andhra Pradesh State Class 10th (SSC) Students can download AP SSC Science model paper 2023 with Answer solutions in Chapter by Chapter for Physical Science, Chemistry, Biological Science and Environmental science for AP SSC Science Model Paper 2023 examination test for Unit Test-1, Unit Test-2, AP SSC General Science Model Paper Unit Test-3, Unit Test-4, and Quarterly, Half Yearly, Pre-Final with Annual final examination tests 2023. The Andhra Pradesh State Board of Secondary Education has published the General Science model paper with study material with practice paper with mock test question bank as AP 10th Biology, PS, Chemistry Model Paper 2023 for all examination tests conducted by BSEAP.


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