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hdu 1258(深搜)

2024-07-21 06:20:09   240人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1258

Sum It Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4012    Accepted Submission(s): 2066


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:  首先给你一个t,一个n;紧接着n个数，问这n个数有多少种加和方式使得加和结果等于t;并且注意不能重复~比如 4=3+1，那么多个4=3+1只能算一种，4=1+3也是一种；

思路：从大到小排序~深搜AC；

 

#include <iostream>
#include <string.h>
#include <string>
#include <stdio.h>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
int t,n,flag;
int a[16],queue[16];//模拟队列

int cmp(int a,int b)
{
    return a>b;
}

void dfs(int sum,int count,int pos)
{
    if(sum>t)return; //递归出口（目标检测函数）
    if(sum==t)
    {
        flag=1;
        for(int i=0;i<count-1;i++)
            printf("%d+",queue[i]);
        printf("%d\n",queue[count-1]);
        return;
    }
    for(int i=pos;i<n;i++)//将一种状态转化为另一种状态的操作结合
    {
        sum+=a[i];
        queue[count]=a[i];
        dfs(sum,count+1,i+1);
        sum-=a[i];
        while(a[i]==a[i+1])i++; //这是为了避免重复
    }
}

int main()
{
        while(cin>>t>>n)
        {
            if(t==0&&n==0)break;
            for(int i=0;i<n;i++)
            {
               cin>>a[i];
            }
            sort(a,a+n,cmp);
            flag=0;
            cout<<"Sums of "<<t<<":"<<endl;
            dfs(0,0,0); //初始状态集合
            if(!flag)
                cout<<"NONE"<<endl;
           
                
        }
        return 0;
}


hdu 1258(深搜)


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