标签:style blog os io strong for ar 2014 div
Consider this sequence {1, 2, 3, … , N}, as a initial sequence of firstN natural numbers. You can rearrange this sequence in many ways. Therewill be N! different arrangements. You have to calculate the number ofarrangement of first N natural numbers, where in first M (M<=N)positions, exactly K (K<=M) numbers are in its initial position.
Example:
For, N = 5, M = 3, K =2
You should count this arrangement {1, 4, 3, 2, 5}, here in first 3positions 1 is in 1st position and 3 in 3rd position. Soexactly 2 of its first 3 are in there initial position.
But you should not count this {1, 2, 3, 4, 5}.
Input
The first line ofinput is an integer T(T<=1000) that indicates the number of testcases. Next T line contains 3 integers each, N(1<=N<=1000), M,and K.
Output
For each case,output the case number, followed by the answer modulo 1000000007. Lookat the sample for clarification.
1 |
Case 1: 12
|
Special Thanks : Abdullah Al Mahmud, Jane Alam Jan
题意:可以把序列1-n任意重排,但重排后的前m个位置中恰好要有k个是不变的,输入整数n,m,k,输出重排数%1000000007.
思路:首先从前m个选出k作为不变的,然后剩下的n-k个再任意重排,也是从中选出i个来作为不变的,剩下的错排。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> typedef long long ll; using namespace std; const int maxn = 1005; const ll mod = 1000000007; ll dp[maxn], C[maxn][maxn]; int n, m, k; void init() { memset(C, 0, sizeof(C)); C[0][0] = 1; for (int i = 1; i < maxn; i++) { C[i][0] = C[i][i] = 1; for (int j = 1; j < i; j++) C[i][j] = (C[i-1][j] + C[i-1][j-1]) % mod; } dp[1] = 0, dp[0] = dp[2] = 1; for (int i = 3; i < maxn; i++) dp[i] = ((i - 1) * (dp[i-2] + dp[i-1]) % mod) % mod; } ll solve() { ll ans = 0; int t = n - m; for (int i = 0; i <= t; i++) ans = (ans + C[t][i] * dp[n-k-i]) % mod; return (ans * C[m][k]) % mod; } int main() { init(); int cas = 1; int t; scanf("%d", &t); while (t--) { scanf("%d%d%d", &n, &m, &k); printf("Case %d: %lld\n", cas++, solve()); } return 0; }
UVA - 11481 Arrange the Numbers
标签:style blog os io strong for ar 2014 div
原文地址:http://blog.csdn.net/u011345136/article/details/38778121