UVA 11077 - Find the Permutations(递推)
2017-04-28 14:05
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UVA 11077 - Find the Permutations
题意:给定n,k求出有多少个包括元素[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个元素须要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,这是能够证明的。原序列中有几个循环,须要的次数就是全部循环长度-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,其余位置都会增加其它循环中,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; }
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