hdu5399Too Simple

来源：互联网发布：mac怎么使用远程桌面编辑：程序博客网时间：2024/06/08 02:23
//给m个函数//其对应是自变量x属于{1,2,...n}//f(x)属于{1,2...3}//给出其中一些函数，问有多少种不同的函数集合使得//1<=i<=n  f1(f2(f3...fm(i))) = i//直接为(m!)^(sum-1)  sum为不知道的函数个数#include<cstdio>#include<cstring>#include<iostream>using namespace std ;const int maxn = 110 ;typedef long long ll ;const ll mod = 1e9+7 ;int map[maxn][maxn] ;int vis[maxn];int a[maxn] ;int main(){    //freopen("in.txt" ,"r" ,stdin) ;    //freopen("out.txt","w" ,stdout);    int n , m ;    int t  = 0 ;    while(~scanf("%d%d" ,&m , &n))    {        int sum = 0 ;        int flag = 0 ;        for(int i = 1;i <= n;i++)        {            memset(vis , 0 , sizeof(vis)) ;            int tmp ;            scanf("%d" , &tmp);            if(tmp == -1)            sum++;            else            {                map[i][1] = tmp;                if(tmp < 1 || tmp > m || vis[tmp])flag = 1;                for(int j = 2;j <= m;j++)                {                    scanf("%d" , &map[i][j]) ;                    if(map[i][j] < 1 || map[i][j] > m || vis[map[i][j]])                    flag = 1;                    vis[map[i][j]] = 1;                }            }        }        if(flag)        {            puts("0");            continue ;        }        if(sum)        {            ll ans = 1;            sum--;            ll tmp = 1 ;            for(ll i = 1;i <= m;i++)            tmp = (tmp*i)%mod ;            while(sum--)            ans = (ans*tmp)%mod ;            printf("%lld\n" , ans) ;            continue ;        }        for(int i = 1;i <= m;i++)        a[i] = i ;        for(int i = n;i > 0;i--)          for(int j = 1;j <= m;j++)          a[j] = map[i][a[j]] ;        flag = 0 ;        for(int i = 1;i <= m;i++)        if(a[i] != i)        {            flag = 1;            break ;        }       if(flag)puts("0");       else puts("1");    }    return 0 ;}
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