zoukankan      html  css  js  c++  java
  • hdu1256

                                      Sum It Up

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2983    Accepted Submission(s): 1494


    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
     
     
    #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){
            printf("%d",res[0]);
            for(int i=1;i<cnt;i++)
                printf("+%d",res[i]);
            printf("
    ");
            flag=1;
        }
        for(int i=cur;i<k;i++){
            res[cnt]=a[i];
            DFS(i+1,sum+a[i],cnt+1);
            while(i+1<k && a[i]==a[i+1])    //去除相同的分解式
                i++;
        }
    }
    
    int main()
    {
        while(scanf("%d%d",&n,&k)!=EOF && n)
        {
            for(int i=0;i<k;i++)
                scanf("%d",&a[i]);
            printf("Sums of %d:
    ",n);
            flag=0;
            DFS(0,0,0);
            if(!flag)
                printf("NONE
    ");
        }
        return 0;
    }
     
     
     
  • 相关阅读:
    铺地毯
    解方程
    引水入城
    10.16今日暂时停更博客
    聪明的质监员
    CCF NOI plus 201(7)6 初赛题 解题报告
    初赛可能会用到的计算机基础理论知识整理
    火柴排队
    借教室
    10.10今日暂时停更博客
  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/3221189.html
Copyright © 2011-2022 走看看