tjut 5399
来源:互联网 发布:企业重要数据要及时 编辑:程序博客网 时间:2024/04/30 03:46
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<set> #include<map> #include<queue> #include<stack> #include<cmath> #include<cstdio> #include<vector> #include<string> #include<cstring> #include<algorithm> #include<functional> using namespace std; typedef long long LL; const LL base = 1e9 + 7; const int maxn = 105; LL T, n, m, f[maxn], a[maxn][maxn]; inline void read(int &x) { char ch; while ((ch = getchar())<'0' || ch>'9'); x = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0'; } int main() { //read(T); for (int i = f[0] = 1; i <= 100; i++) f[i] = f[i - 1] * i % base; while (scanf("%I64d%I64d", &n, &m) != EOF) { LL tot = 0, ans = 1; for (int i = 1; i <= m; i++) { scanf("%I64d", &a[i][1]); if (a[i][1] == -1) tot++; else for (int j = 2; j <= n; j++) { scanf("%I64d", &a[i][j]); for (int k = j - 1; k; k--) if (a[i][j] == a[i][k]) ans = 0; } } for (int i = 1; i < tot; i++) ans = ans * f[n] % base; if (tot == 0) { for (int i = 1; i <= n; i++) a[0][i] = i; for (int i = m; i; i--) for (int j = 1; j <= n; j++) a[0][j] = a[i][a[0][j]]; for (int i = 1; i <= n; i++) if (a[0][i] != i) ans = 0; } printf("%I64d\n", ans); } return 0; }
0 0
- tjut 5399
- tjut 5289
- tjut 5288
- tjut 5294
- tjut 2586
- tjut 5296
- tjut 5297
- tjut 5299
- tjut 5384
- tjut 5387
- tjut 5386
- tjut 5381
- tjut 5400
- tjut 5396
- tjut 5398
- tjut 5412
- tjut 5410
- tjut 5416
- submit表单提交
- css中的外边距合并时垂直方向上的普通流相邻元素间
- APP开发实战110-ProGuard简介
- Java-I/O
- Java下利用Jackson进行JSON解析和序列化
- tjut 5399
- JavaSE基础(二)
- python3提示sqlite3模块不存在解决方法
- 概率,递推,找规律,高精度(FXTZ II,hdu 4043)
- GDB 调试 C++ 程序 core dump
- 快速排序
- JVM运行时数据区
- ubuntu 安装截图工具 Shutter,并设置快捷键 Ctrl+Alt+A
- 日常使用javascript和JQuery函数总结