Sum It Up NSOJ

最新推荐文章于 2021-05-24 19:02:03 发布

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本文介绍了一种解决特定求和问题的算法方法，通过详细解析一个实例问题，即找出列表中所有可能组合来达到预定的目标和。文章深入探讨了算法实现细节，包括数据预处理、避免重复解的方法及搜索策略。

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Sum It Up

时间限制: 1000ms

内存限制: 32768KB

HDU ID: 1258
64位整型: Java 类名:

上一题

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题目描述

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.

输入

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.

输出

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.

样例输入

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

样例输出

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

来源

浙江工业大学第四届大学生程序设计竞赛

思路：主要是那些搜索前数据的处理，还有就是判断是否重复，都是因为数据的预处理。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#include<map>
int n,m,flag;
int a[1005],b[1005],c[1005];
int book[12];
int cmp(int t1,int t2)
{
    return t1>t2;
}
void  check(int j)
{
    int i;
    for( i=0; i<j; i++)
        if(b[i]!=c[i])
         break;
    if(i<j)
    {
        for(i=0; i<j; i++)
        {
            c[i]=b[i];
            if(i==0)
                printf("%d",b[i]);
            else printf("+%d",b[i]);
        }
        printf("\n");
    }
    return ;
}
void dfs(int sum,int step,int k)
{
    if(sum==m)
    {
        check(step);
        flag=1;
        return;
    }
    for(int i=k+1; i<n; i++)
        if(!book[i]&&a[i]+sum<=m)
        {
            b[step]=a[i];
            book[i]=1;
            dfs(sum+a[i],step+1,i);
            book[i]=0;
            while(a[i]==a[i+1]&&i<=n)
                i++;
        }
    return ;
}
int main()
{
    while(~scanf("%d %d",&m,&n)&&(n+m))
    {
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n,cmp);
        flag=0;
        printf("Sums of %d:\n",m);
        memset(c,0,sizeof(c));
        for(int i=0; i<n; i++)
        {
            memset(book,0,sizeof(book));
            if(a[i]>m)
                continue;
            if(i>0&&a[i]!=a[i-1])
            {
                book[i]=1;
                b[0]=a[i];
                dfs(a[i],1,i);
            }
            if(i==0)
            {
                book[i]=1;
                b[0]=a[0];
                dfs(a[i],1,i);
            }
        }
        if(!flag)
            printf("NONE\n");
    }
    return 0;
}