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  • GCPC2014 A Algebraic Teamwork

    题意:其实就是要你求排列数。

    解题思路:(n!-1)%1000000009

    解题代码:

     1 #include <iostream>
     2 #include <string.h>
     3 #include <stdio.h>
     4 
     5 using namespace std;
     6 typedef long long LL;
     7 #define M 1000000007
     8 #define maxn 100005
     9 int n,m,p;
    10 LL c[maxn];
    11 LL dp[maxn];
    12 LL quick_mod(LL a, LL b)
    13 {
    14     LL ans = 1;
    15     a %= p;
    16     while(b)
    17     {
    18         if(b & 1)
    19         {
    20             ans = ans * a % p;
    21             b--;
    22         }
    23         b >>= 1;
    24         a = a * a % p;
    25     }
    26     return ans;
    27 }
    28 LL C(int n)
    29 {
    30     LL ans = 1;
    31     for(int i=1; i<= n; i++)
    32     {
    33         LL a = (n+1-i);
    34         LL b = i ;
    35         ans = ans * (a * quick_mod(b, p-2) % p) % p;
    36         c[i] = ans ;
    37     }
    38     return ans;
    39 }
    40 int main()
    41 {
    42     int T;
    43     scanf("%d", &T);
    44     dp[1] = 1;
    45     c[0] = 1;
    46     dp[1] = 1;
    47     for(int i = 2;i <= 100000;i ++ )
    48         dp[i] = (i * dp[i-1])%M;
    49     while(T--)
    50     {
    51         p = M ; 
    52         scanf("%d",&n);
    53         LL sum ; 
    54         sum = (dp[n]-1)%M;
    55         printf("%lld
    ",sum);
    56     }
    57     return 0;
    58 }
    View Code
    没有梦想,何谈远方
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  • 原文地址:https://www.cnblogs.com/zyue/p/4340488.html
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