UVA 11077 - Find the Permutations(递推)
2014-07-18 12:56
591 查看
UVA 11077 - Find the Permutations
题目链接题意:给定n,k求出有多少个包含元素[1-n]的序列,交换k次能得到一个[1,2,3...n]的序列
思路:递推dp[i][j]表示i个元素需要j次,那么在新加一个元素的时候,添在最后面次数不变,其余位置都是次数+1,这是可以证明的,原序列中有几个循环,需要的次数就是所有循环长度-1的和,那么对于新加一个元素,加在最后就和自己形成一个循环,次数不变,其余位置都会加入其他循环中,次数+1,因此递推式为dp(i,j)=dp(i−1,j−1)∗(i−1)+dp(i−1,j)
代码:
#include <stdio.h>
#include <string.h>
const int N = 22;
int n, k;
unsigned long long dp
;
int main() {
dp[1][0] = 1;
for (unsigned long long i = 2; i <= 21; i++) {
for (int j = 0; j < i; j++) {
dp[i][j] = dp[i - 1][j];
if (j) dp[i][j] += dp[i - 1][j - 1] * (i - 1);
}
}
while (~scanf("%d%d", &n, &k) && n || k) {
printf("%llu\n", dp
[k]);
}
return 0;
}
相关文章推荐
- UVA 11077 Find the Permutations 置换+递推
- UVa 11077 Find the Permutations(置换+递推)
- UVA 11077 - Find the Permutations(递推)
- uva 11077 Find the Permutations 置换+递推
- UVA 11077 Find the Permutations(置换)
- uva 11077 - Find the Permutations(置换+dp)
- Uva 11077 Find the Permutations [置换群 DP]
- UVA 11077 Find the Permutations(置换+dp)
- UVA 11077 Find the Permutations 递推置换
- UVA 11077 Find the Permutations DP
- UVA11077 Find the Permutations
- uva11077 Find the Permutations
- UVA - 11077 Find the Permutations (置换)
- 【UVA 11077】 Find the Permutations (置换+第一类斯特林数)
- UVa11077-Find the Permutations(dp+置换)
- UVA - 11077 Find the Permutations (置换)
- UVa 11077 Find the Permutations / 置换
- UVA 11077 Find the Permutations
- uva 11077 - Find the Permutations(置换)
- UVa 11077 (循环分解 递推) Find the Permutations