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  • HDU5399-多校-模拟

    Too Simple

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 1214    Accepted Submission(s): 406


    Problem Description
    Rhason Cheung had a simple problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.

    Teacher Mai has m functions f1,f2,,fm:{1,2,,n}{1,2,,n}(that means for all x{1,2,,n},f(x){1,2,,n}). But Rhason only knows some of these functions, and others are unknown.

    She wants to know how many different function series f1,f2,,fm there are that for every i(1in),f1(f2(fm(i)))=i. Two function series f1,f2,,fm and g1,g2,,gm are considered different if and only if there exist i(1im),j(1jn),fi(j)gi(j).
     
    Input
    For each test case, the first lines contains two numbers n,m(1n,m100).

    The following are m lines. In i-th line, there is one number 1 or n space-separated numbers.

    If there is only one number 1, the function fi is unknown. Otherwise the j-th number in the i-th line means fi(j).
     
    Output
    For each test case print the answer modulo 109+7.
     
    Sample Input
    3 3 1 2 3 -1 3 2 1
     
    Sample Output
    1
    Hint
    The order in the function series is determined. What she can do is to assign the values to the unknown functions.
     
     
    一开始看起来很复杂,想了想发现只要保证最后一个函数是任意的,中间的其他函数都没问题。
    但是有一个坑的情况是,可能一个任意函数也没有,全都是固定的函数,这时要验证这组函数是否可行。我们当时验证的顺序弄反,卡了一场,真是too simple。
    最近做的题都是模拟题啊,,,太没技术含量了,,,,
     
     1 #include <cstdio>
     2 #include <algorithm>
     3 #include <cstring>
     4 #include <ctype.h>
     5 #include <cstdlib>
     6 #include <stack>
     7 #include <set>
     8 #include <map>
     9 #include <queue>
    10 #include <string>
    11 #include <cmath>
    12 
    13 using namespace std;
    14 const long long MOD = 1e9+7;
    15 
    16 int N,T,M;
    17 int func[200][200],vis[200];
    18 long long nn;
    19 
    20 long long qpow(long long a,long long i,long long n)
    21 {
    22     if(i == 0) return 1 % n;
    23     long long temp = qpow(a,i>>1,n);
    24         temp = temp * temp % n;
    25     if( i&1 ) temp = temp * a % n;
    26     return temp;
    27 }
    28 
    29 long long mi(long long a,int t)
    30 {
    31     long long ans = 1;
    32     for(int i=0;i<t;i++) {ans *= a;ans %= MOD;}
    33     return ans;
    34 }
    35 
    36 long long solve()
    37 {
    38     for(int i=1;i<=N;i++)
    39     {
    40         int ans = i;
    41         for(int j=M-1;j>=0;j--)
    42         {
    43             ans = func[j][ans-1];
    44         }
    45         if(ans != i) return 0LL;
    46     }
    47     return 1LL;
    48 }
    49 
    50 int main()
    51 {
    52     while(~scanf("%d%d",&N,&M))
    53     {
    54         long long cnt = 0LL,ans = 0LL;
    55         nn = 1LL;
    56         int flag = 0;
    57         for(int i=1; i <= N;i++) {nn *= i; nn %= MOD;}
    58 
    59         for(int i=0;i<M;i++)
    60         {
    61             memset(vis,0,sizeof vis);
    62             if(scanf("%d",&func[i][0]) && (func[i][0] == -1))
    63             {
    64                 cnt++;
    65             }
    66             else 
    67             {
    68                 vis[func[i][0]]++;
    69                 for(int j=1;j<N;j++) 
    70                 {
    71                     scanf("%d",&func[i][j]);
    72                     if( vis[func[i][j]] ) flag = 1;
    73                     else vis[func[i][j]]++;
    74                 }
    75             }
    76         }
    77 
    78         if(flag) ans = 0LL;
    79         else if(cnt > 0) { ans = mi(nn,cnt-1); ans %= MOD;}
    80         else
    81         {
    82             ans = solve();
    83         }
    84 
    85         printf("%I64d
    ",ans%MOD);
    86     }
    87 }
     
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  • 原文地址:https://www.cnblogs.com/helica/p/4783618.html
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