标签：des style http os io for ar art
Description
Tamim and Lina, two of the biggest mega minds of Bangladesh went to a far country. They ate, coded and wandered around, even in their holidays. They passed several months in this way. But everything has an end. A holy person, e-Moti-ji came into their life. e-Moti-ji took them to derby (horse racing). e-Moti-ji enjoyed the race, but as usual Tamim and Lina did their as usual task instead of passing some romantic moments. They were thinking- in how many ways a race can finish! Who knows, maybe this is their romance!
In a race there are n horses. You have to output the number of ways the race can finish. Note that, more than one horse may get the same position. For example, 2 horses can finish in 3 ways.
Input
Input starts with an integer T (
1000), denoting the number of test
cases. Each case starts with a line containing an integer n (1n1000).
Output
For each case, print the case number and the number of ways the race can finish. The result can be very large, print the result modulo 10056.
Sample Input
3 1 2 3
Sample Output
Case 1: 1 Case 2: 3 Case 3: 13题意：求n匹马可能的排列情况（可以并列）
思路：设dp[i][j]表示i匹马j个排名的情况，那么我们可以发现在dp[i-1][j]和dp[i-1][j-1]都可以在有j个插入可能
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1005;
const int mod = 10056;
int n, dp[maxn][maxn];
int ans[maxn];
void init() {
dp[0][0] = 1;
for (int i = 1; i < maxn; i++) {
int sum = 0;
for (int j = 1; j <= i; j++) {
dp[i][j] = (dp[i-1][j-1] + dp[i-1][j]) * j % mod;
sum = (sum + dp[i][j]) % mod;
}
ans[i] = sum;
}
}
int main() {
int t, cas = 1;
init();
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
printf("Case %d: %d\n", cas++, ans[n]);
}
return 0;
}
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标签：des style http os io for ar art
原文地址：http://blog.csdn.net/u011345136/article/details/38583121
