Sum It Up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 213 Accepted Submission(s): 106

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 15 3 2 1 1400 12 50 50 50 50 50 50 25 25 25 25 25 250 0

 

Sample Output

Sums of 4:43+12+22+1+1Sums of 5:NONESums of 400:50+50+50+50+50+50+25+25+25+2550+50+50+50+50+25+25+25+25+25+25

题目大意：给出一个数字n，再给出k个数字，在这些数字中选取数字相加的和等于n（选几个数不限，一个式子中数字的使用次数不能超过给出的数字的个数）

思路：输出要求从大到小，数据量很小，暴力搜索，搜索时注意当前搜索的起点不能和上一次一样，避免重复。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,k,a[1010],flag,res[1010];void DFS(int cur,int sum,int cnt){    if(sum>n)        return ;    if(sum==n){        cout<<res[0];        for(int i=1;i<cnt;i++)            cout<<"+"<<res[i];        cout<<endl;        flag=1;    }    for(int i=cur;i<k;i++){        res[cnt]=a[i];        DFS(i+1,sum+a[i],cnt+1);        while(a[i]==a[i+1]&&i+1<k)//和之前一样  跳过本次搜索            i++;    }}int main(){    while(cin>> n>>k){        if(n==0)break;        for(int i=0;i<k;i++)            cin>>a[i];        cout<<"Sums of "<<n<<":"<<endl;        flag=0;        DFS(0,0,0);        if(!flag)            cout<<"NONE"<<endl;    }    return 0;}