标签：std class zoj \n size can $$ mes lld
题目大意：给出两个$n\times n$的矩阵，$m$次询问它们的积中给定子矩阵的数值和。
题解：令为$P\times Q=R$
$$\begin{align*}
&\sum\limits_{i=a}^c\sum\limits_{j=b}^dR[i][j]\\
=&\sum\limits_{i=a}^c\sum\limits_{j=b}^d\sum\limits_{k=1}^nP[i][k]\cdot Q[k][j]\\
=&\sum\limits_{k=1}^n(\sum\limits_{i=a}^cP[i][k])\cdot (\sum\limits_{j=b}^dQ[k][j])\\
=&\sum\limits_{k=1}^n(sumP[c][k]-sumP[a-1][k])(sumQ[k][d]-sumQ[k][b-1])\\
\end{align*}$$
卡点：无
C++ Code：
#include <cstdio>
#define int long long
#define maxn 2005
using namespace std;
int n, Q, a, b, c, d, ans;
int p[maxn][maxn], q[maxn][maxn];
int sp[maxn][maxn], sq[maxn][maxn];
inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
signed main() {
scanf("%lld%lld", &n, &Q);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) scanf("%lld", &p[i][j]), sp[i][j] = sp[i - 1][j] + p[i][j];
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) scanf("%lld", &q[i][j]), sq[i][j] = sq[i][j - 1] + q[i][j];
}
while (Q --> 0) {
scanf("%lld%lld%lld%lld", &a, &b, &c, &d);
if (a > c) swap(a, c);
if (b > d) swap(b, d);
ans = 0;
for (int i = 1; i <= n; i++) ans += (sp[c][i] - sp[a - 1][i]) * (sq[i][d] - sq[i][b - 1]);
printf("%lld\n", ans);
}
return 0;
}
标签：std class zoj \n size can $$ mes lld
原文地址：https://www.cnblogs.com/Memory-of-winter/p/9497641.html
