Sicily 1715. A piece of cake

最新推荐文章于 2016-06-09 00:00:23 发布

原创最新推荐文章于 2016-06-09 00:00:23 发布 · 441 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#sicily

sicily 专栏收录该内容

413 篇文章

订阅专栏

本文介绍了一个简单的计数问题：如何计算将n个不同的球放入m个盒子中，每个盒子至少包含k个球的不同方式的数量。通过使用组合数学的方法，文章提供了一种有效的算法实现方案。

1715. A piece of cake

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

This problem is very simple, i.e. just a piece of cake for you, excellent programmers. You just need to calculate and output how many ways to put n different balls into m different boxes so that each box has at least k balls.

Input

Input contains several test cases. Each of the test cases contains three integers in one line, n, m, k (1 <= n, m <= 15, 0 <= k <= 15). Input is terminated by three 0s, which should not be processed.

Output

For each case, just print the result in one line. Heading zeros are forbidden. For example, 12 is legal output but 012 is not, 0 is legal but 00 is not, and so on.

Sample Input

Sample Output

6
0
8
// Problem#: 1715
// Submission#: 3585085
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <stdio.h>
int n, m, k;

long long mat[60][60];

long long gcd(long long a, long long b) {
    long long r;
    while (b) {
        r = a % b;
        a = b;
        b = r;
    }
    return a;
}

long long c(int a, int b) {
    long long x = 1, y = 1;
    long long r;
    int i;
    for (i = 0; i < b; i++) {
        x *= (a - i);
        y *= (b - i);
        r = gcd(x, y);
        x /= r;
        y /= r;
    }
    return x / y;
}

void process() {
    int i, j, u;
    for (i = 0; i <= n; i++)
        for (j = 0; j <= m; j++) mat[i][j] = 0;
    for (i = 0; i <= n; i++) mat[i][0] = 0;
    mat[0][0] = 1;
    for (i = 1; i <= m; i++)
        for (j = k * i; j <= n; j++) {
            mat[j][i] = 0;
            for (u = k; u <= j - (i - 1) * k; u++) {
                long long temp = c(j, u);
                mat[j][i] = mat[j][i] + temp * mat[j - u][i - 1];
            }
        }
    printf("%lld\n", mat[n][m]);
}

int main() {
    while (scanf("%d%d%d", &n, &m, &k) == 3) {
        if (n == 0) break;
        process();
    }
    return 0;
}