POJ 1564 Sum It Up（深搜）

Description:

Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n = 6, and the list is [4, 3, 2, 2, 1, 1], then there are four different sums that equal 4: 4, 3+1, 2+2, and 2+1+1. (A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.

Input:

The input will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers x 1 , . . . , x n . If n = 0 it signals the end of the input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12 (inclusive), and x 1 , . . . , x n will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear in nonincreasing order, and there may be repetitions.

Output:

For each test case, first output a line containing `Sums of', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line `NONE'. The numbers within each sum must appear in nonincreasing order. A number may be repeated in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distinct; the same sum cannot appear twice.

Sample Input:

4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0

Sample Output:

Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25

题意：有n个数，在这n个数中找到和是sum的数，将这些数按a+b+c+...的格式输出，答案可能不止一个，也可能没有，输入时数组不是降序（包括重复的元素）输入的，输出时也是降序（包括重复）。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int N=110;
int a[N], b[N], sum, n, flag;

void DFS(int idex, int k, int s)
{
    int i;

    if (s == sum) ///和与sum相等了，则输出这个序列
    {
        flag = 1; ///标记能否找到这样的序列
        printf("%d", b[0]);
        for (i = 1; i < k; i++)
            printf("+%d", b[i]);
        printf("
");

        return ;
    }
    else
    {
        for (i = idex; i < n; i++)
        {
            if (s + a[i] <= sum) ///只有相加小于sum，才有可能等于sum
            {
                b[k] = a[i];
                DFS(i+1, k+1, s+a[i]);
                while (i+1 < n && a[i+1] == a[i]) i++; ///答案不能有重复的，那么重复的元素要跳过
            }
        }
    }
}
int main ()
{
    int i;

    while (scanf("%d%d", &sum, &n), n)
    {
        for (i = 0; i < n; i++)
            scanf("%d", &a[i]);

        printf("Sums of %d:
", sum);
        flag = 0;
        DFS(0, 0, 0);
        if (flag == 0) printf("NONE
");
    }

    return 0;
}