UVA 12034 - Race(递推)
2014-05-23 19:47
239 查看
UVA 12034 - Race
题目链接题意:给定n匹马,要求出可能的排名情况(可能并列)
思路:递推,dp[i][j]表示i匹马的时候有j种不同名次,那么dp[i][j]可以由dp[i - 1][j - 1]插入j个不同位置得来,或者由dp[i - 1][j]放入已有j的名次得来,得到递推式dp[i][j] = j * (dp[i - 1][j - 1] + dp[i - 1][j]); 然后对于n的答案为sum{dp
[j]} (1 <= j <= n)
代码:
#include <stdio.h> #include <string.h> const int MOD = 10056; const int N = 1005; int t, n, dp , ans ; void init() { dp[0][0] = 1; for (int i = 1; i <= 1000; i++) { int sum = 0; for (int j = 1; j <= i; j++) { dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % MOD * j % MOD; sum = (sum + dp[i][j]) % MOD; } ans[i] = sum; } } int main() { int cas = 0; init(); scanf("%d", &t); while (t--) { scanf("%d", &n); printf("Case %d: %d\n", ++cas, ans ); } return 0; }
相关文章推荐
- 【UVA】12034-Race(递推,组合数打表)
- CSU-ACM2017暑期训练3-递推与递归 H - Race UVA - 12034
- UVA 12034 Race(递推)
- UVa 12034 (递推) Race
- UVa 12034 - Race(组合+递推)
- Race - UVa 12034 递推
- uva 12034 Race递推
- UVa12034 Race 递推
- UVA 12034 Race 动态规划+递推
- uva12034(递推关系)
- UVA 12034 Race
- UVa 12034 比赛名次(递推)
- uva 12034 Race(dp)
- uva 12034 - Race(dp计数)
- UVA 12034(p332)----Race
- UVA 12034 Race
- UVA 12034 Race 赛马名次
- (UVA - 12034)Race(组合数,dp)
- UVA 12034 Race ——dp
- uva 12034 Race(递推+组合数)