hdu 1258 Sum It Up(dfs)

Sum It Up

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

Total Submission(s): 4731    Accepted Submission(s): 2461

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

分析：非单调递减顺序输出答案；

code：

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;int n,t;int f[20],a[20],aa[20],ans,tt,cot;int cmp(int a,int b){    return a>b;}void dfs(int num){    if(ans==t)    {        cot++;        printf("%d",aa[0]);//aa数组存放满足条件的组合，        for(int i=1;i<tt;i++)            printf("+%d",aa[i]);        printf("\n");        return ;    }    if(ans>t)        return ;    for(int k=num;k<n;k++){        if(!f[k]){            aa[tt++]=a[k];            f[k]=1;            ans+=a[k];            dfs(k+1);            f[k]=0;//回溯            ans-=a[k];            tt--;        }        while(k+1<n&&a[k]==a[k+1])//与当前遍历数字相同的数字应该直接跳过 即去重，因为在深搜过程中下一个遍历的数 在下一层递归循环中，这样允许相同//的数字多次出现时可能全部被放入满足条件的数组中 且保证了不会有相同的答案出现。            k++;    }}int main(){    int flag,tot,minn;    while(scanf("%d%d",&t,&n)){        if(t==0&&n==0)            break;        fill(f,f+20,0);        for(int i=0; i<n; i++)            scanf("%d",&a[i]);                    cot=0;ans=0;tt=0;        sort(a,a+n,cmp);        printf("Sums of %d:\n",t);        dfs(0);        if(!cot)            cout<<"NONE"<<endl;    }    return 0;}