m-ary Partitions

最新推荐文章于 2020-02-19 19:31:26 发布

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最新推荐文章于 2020-02-19 19:31:26 发布

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分类专栏： spiny 文章标签：动态规划 dp

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一、问题：

Description

A partition of an integer n is a set of positive integers which sum to n, typically written in descending order. For example: 10 = 4+3+2+1 A partition is m-ary if each term in the partition is a power of m. For example, the 3-ary partitions of 9 are:

3+3+3

3+3+1+1+1

3+1+1+1+1+1+1 1+1+1+1+1+1+1+1+1

Write a program to find the number of m-ary partitions of an integer n.

Input

The first line of input contains a single decimal integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set should be processed identically and independently. Each data set consists of a single line of input. The line contains the data set number, K, followed by the base of powers, m, (3 ≤ m ≤ 100), followed by a space, followed by the integer, n, (3 ≤ n ≤ 10000), for which the number of m-ary partitions is to be found.

Output

For each data set there is one line of output. The output line contains the data set number, K, a space, and the number of m-ary partitions of n. The result should fit in a 32-bit unsigned integer.

Sample Input

1 3 9

2 3 47

3 5 123

4 7 4321

5 97 9999

Sample Output

1 5

2 63

3 75

4 144236

5 111

二、题意：

第一行数字 T 代表测试样例个数。

接下来的 T 行每行第一个数 cnt 表示序号，第二个数m，第三个数n。

求：使不超过n的m的幂(每种幂有无穷个)的和等于n，有多少种组合方法。注意：幂的种类和各种幂的个数均相同的视为同一种，如3+1+1+1和1+1+1+3为同一种。

三、思路：

对于每一个数n，都可以拆成两个数的和，那么拼出来n的方法数就等于 (拼出n-1的方法+拼出1的方法) + (拼出n-2的方法+拼出2的方法) + ..... 下面代码中的 dp数组记录的就是拼出n的方法数。

四、代码：

#include <cstdio>
#include <cstring>
#define read(x) scanf("%d",&x)
#define mem(a,b) memset(a,0,sizeof(a))
#define fori(a,b) for(int i=a;i<=b;i++)
#define forj(a,b) for(int j=a;j<=b;j++)

int main()
{
    int T;
    read(T);
    while(T--)
    {
        int dp[11000], a[110];
        int cnt, m, n, z=1, k=-1;
        scanf("%d %d %d", &cnt, &m, &n);
        mem(a, 0);
        mem(dp, 0);
        dp[0] = 1;
        while(z <= n)
        {
            a[++k] = z;
            z *= m;
        }
        fori(0, k)
            forj(a[i], n)
                dp[j] += dp[j-a[i]];
        printf("%d %d\n", cnt, dp[n]);
    }
    return 0;
}