HDU 1258(dfs)

题意：给出t，n，还有n个非递增的正整数，求用这些非递增的正整数中挑选出可以相加成t的组合。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;

int num[101];
int amount[101];
int box[13];
bool done;
void DFS(int sum, int end, int sub, int start)
{
    if (sub > 12)
        return;
    if (sum == 0)
    {
        done = true;
        printf("%d", box[1]);
        for (int j = 2; j < sub; ++j)
            printf("+%d", box[j]);
        printf("\n");
        return;
    }
    for (int j = start; j <= end; ++j)
    {
        if (amount[j])
        {
            if (sum >= num[j])
            {
                --amount[j];
                box[sub] = num[j];
                DFS(sum-num[j], end, sub+1, j);
                ++amount[j];
            }
        }
    }
}

int main()
{
    int t, n, i, tmp, s;
    num[0] = 0;
    while (scanf("%d%d", &t, &n) != EOF && t + n)
    {
        s = i = 0;
        done = false;
        memset(amount, 0, sizeof amount);
        while (n--)
        {
            scanf("%d", &tmp);
            if (tmp > t) continue;
            s += tmp;
            if (tmp != num[i])
            {
                num[++i] = tmp;
                ++amount[i];
            }
            else ++amount[i];
        }
        printf("Sums of %d:\n", t);
        if (i == 0 || s < t)
        {
            printf("NONE\n");
            continue;
        }
        DFS(t, i, 1, 1);
        if (!done) printf("NONE\n");
    }
    return 0;
}