内存限制:256 MiB 时间限制:2000 ms 标准输入输出
题目类型:传统 评测方式:文本比较
上传者: 匿名

模版
以前一直不过样例原来是读入优化没写负数。。
#include <cstdio> #include <cctype> const int Mod = 1e9 + 7; typedef long long LL; #define N 505 #define rep(a,b,c) for (int i = a; i <= b; i+=c) #define Rep(a,b,c) for (int j = a; j <= b; j+=c) #define REP(a,b,c) for (int k = a; k <= b; k+=c) inline void read(LL &x) { bool f = 0;register char ch = getchar(); for (x = 0; !isdigit(ch); ch = getchar()) if (ch == '-') f = 1; for (; isdigit(ch); x = x * 10 + ch - '0', ch = getchar()); x = f ? (-x) : x; } int n, p, m; LL A[N][N], B[N][N], C[N][N]; int Main() { scanf("%d%d%d", &n, &p, &m); rep(1,n,1) Rep(1,p,1) read(A[i][j]); rep(1,p,1) Rep(1,m,1) read(B[i][j]); rep(1,n,1) Rep(1,m,1) { REP(1,p,1) C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % Mod; } rep(1,n,1) Rep(1,m,1) j == m ? printf("%lld ",(C[i][j] + Mod) % Mod) : printf("%lld ",(C[i][j] + Mod) % Mod); return 0; } int sb = Main(); int main(int argc, char *argv[]) {;}