## POJ 1564 Sum It Up（DFS）

2018年01月12日 ⁄ 综合 ⁄ 共 2594字 ⁄ 字号 评论关闭
Sum It Up
 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5509 Accepted: 2778

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
```

（具体的解释写在代码里面吧）

```#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 1000
#define INF 1<<25
using namespace std;
int t,n,a[15]={0},b[15]={0};
bool flag=false;
void dfs(int sum, int x, int p)
{
int i;
if (sum==0)//如果可以组合为sum,输出
{
flag=true;
printf("%d",b[i]);
for (i=1; i<p; i++)
printf("+%d",b[i]);
printf("\n");
return ;
}
for (i=x; i<n; i++)//x代表了是第几层
if (i==x || a[i]!=a[i-1])//如果i=x，即使该层的首个元素，那么肯定可以选，其次是用a[i]!=a[i-1]这个判断删去在同一层中重复的元素
{
b[p]=a[i];
dfs(sum-a[i],i+1,p+1);
}
}
int main ()
{
while(cin>>t>>n)
{
if (n==0) break;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for (int i=0; i<n; i++) cin>>a[i];
flag=false;
printf("Sums of %d:\n",t);
dfs(t,0,0);
if (!flag) printf("NONE\n");
}
return 0;
}
```

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