从给定的N个正数中选取若干个数之和为M

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#include <iostream>
#include <list>
using namespace std;

void find_seq(int sum, int index, int * value, list<int> & seq)
{
    if(sum <= 0 || index < 0) return;
    if(sum == value[index])
    {
        printf("%d ", value[index]);
        for(list<int>::iterator iter = seq.begin(); iter != seq.end(); ++iter)
        {
            printf("%d ", *iter);
        }
        printf("\n");
    }
    seq.push_back(value[index]); 
    find_seq(sum-value[index], index-1, value, seq); //放value[index]
    seq.pop_back();
    find_seq(sum, index-1, value, seq); //不放value[index]
}

int main()
{
    int M;
    list<int> seq;
    int value[] = {2,9,5,7,4,11,10};
    int N = sizeof(value)/sizeof(value[0]);
    for(int i = 0; i < N; ++i)
    {
        printf("%d ",value[i]);
    }
    printf("\n");
    scanf("%d", &M);
    printf("可能的序列：\n");
    find_seq(M, N-1, value, seq);

    return 0;
}

从给定的N个正数中选取若干个数之和为M

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原文地址：http://blog.csdn.net/u014082714/article/details/43371141