uva 12034 - Race(dp计数)
2014-05-25 23:20
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题目链接:uva
12034 - Race
题目大意:有n匹马比赛,问说有多少种排名情况,可以并列。解题思路:dp[i][j]表示i匹马,最后一名为j的情况,转移方程dp[i][j]=(dp[i−1][j]+dp[i−1][j−1])∗j
#include <cstdio> #include <cstring> typedef long long ll; const int N = 1005; const ll MOD = 10056; ll dp , f ; void init () { memset(f, 0, sizeof(f)); memset(dp, 0, sizeof(dp)); dp[1][1] = 1; for (ll i = 1; i <= 1000; i++) { for (ll j = 1; j <= i; j++) { dp[i+1][j] = (dp[i+1][j] + dp[i][j] * j) % MOD; dp[i+1][j+1] = (dp[i+1][j+1] + dp[i][j] * (j + 1)) % MOD; f[i] = (f[i] + dp[i][j]) % MOD; } } } int main () { init(); int cas, n; scanf("%d", &cas); for (int i = 1; i <= cas; i++) { scanf("%d", &n); printf("Case %d: %lld\n", i, f ); } return 0; }
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