Sum It Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem 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 x1,...,xn. 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 x1,...,xn 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 distince; the same sum connot 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

题意没什么特别的，非常easy也就看懂了。。也就不多说了。

难点：

这题要推断是否反复，反复的式子仅仅出现一次

也就要保证DFS開始的起点不一样。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

const int M = 1000 + 5;
int n, m;
int num[M];      //数据输入
int ans[M];     //数据输出
int vist;      //标记
int step;
bool flag;

void dfs(int x, int cnt)
{
    if(cnt>n)
        return ;
    else if(cnt==n)      //符合要求则输出
    {
        flag=true;
        printf("%d", ans[1]);
        for(int i=2; i<step; i++)
            printf("+%d", ans[i]);
        printf("\n");
        return ;
    }
    int vist=-1;      //用于推断DFS開始的起点是否同样
    for(int j=x+1; j<=m; j++)
    {
        if(num[j]!=vist)
        {
            ans[step++]=num[j];
            vist=num[j];
            dfs(j, cnt+num[j]);
            step--;
        }
    }
}

int main()
{
    while(scanf("%d%d", &n, &m) &&n &&m)
    {
        flag=false;
        step=1;
        memset(num, 0, sizeof(num));     //数据初始化
        memset(ans, 0, sizeof(ans));
        for(int i=1; i<=m; i++)
            scanf("%d", &num[i]);
        printf("Sums of %d:\n", n);
        dfs( 0, 0 );
        if(!flag)
            printf("NONE\n");
    }

    return 0;
}

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