UVA12034 Race

嘟嘟嘟

令dp[i]表示在n个人中，有 i 个人获得第一名的方案数，转移方程为dp[i] = C(i, n) * dp[n - i]。C(i, n)就是从n个人中选 i 个第一，那么剩下的n - i 个人必须都不是第一，所以就从dp[n - i]转移过来。

因为模数不是质数，所以O(n²)杨辉三角递推。然后O(n²)dp预处理，O(1)查询。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<stack>
 #include<queue>
 #include<vector>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e3 + ;
 const int mod = ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), las = ' ';
   while(!isdigit(ch)) las = ch, ch = getchar();
   while(isdigit(ch)) ans = ans *  + ch - '', ch = getchar();
   if(las == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) putchar('-'), x = -x;
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int c[maxn][maxn], dp[maxn];

 int main()
 {
   c[][] = ;
   for(int i = ; i < maxn; ++i)
     {
       c[i][] = ;
       for(int j = ; j < maxn; ++j)
     c[i][j] = (c[i - ][j] + c[i - ][j - ]) % mod;
     }
   dp[] = ;
   for(int i = ; i < maxn; ++i)
       for(int j = ; j <= i; ++j)
     dp[i] = (dp[i] + c[i][j] * dp[i - j] % mod) % mod;
   int T = read();
   for(int i = ; i <= T; ++i)
     {
       int n = read();
       printf("Case %d: %d\n", i, dp[n]);
     }
   return ;
 }