POJ 1564 Sum It Up

题目链接：http://poj.org/problem?id=1564

 

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 x 1 , . . . , x n . 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 x 1 , . . . , x n 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 distinct; the same sum cannot 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

想了半个多小时，打了一个很挫的代码，还不对。甚是无语！%>_<%

后来看了一下别人的代码，发现判重判的很犀利，短短的代码涵盖了所有要求，牛啊！

最主要的代码：（即判重）

for(int i=index;i<=n;++i)
    {
        if((i==index || num[i]!=num[i-1])&&(num[i]<=left))
           {
               step[cnt]=num[i];
               dfs(i+1,cnt+1,left-num[i]);
           }
    }

 我在看了别人的代码以后自己编了一个，又加了一些优化：

1.输入的数比t大的时候可以去掉

2.要用的数比剩下的数还大时可以跳过

成功优化到0ms

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int step[15];
int num[15];
int t,n;
bool No_Found=1;  //NOT  FOUND ANS
void dfs(int index,int cnt,int left)
{
    if(left==0)
    {
        No_Found=0;
        for(int i=1;i<cnt;++i)
        {
            if(i==1)
            {
                printf("%d",step[i]);
            }
            else
            {
                printf("+%d",step[i]);
            }
        }
        printf("\n");
        return ;
    }
    for(int i=index;i<=n;++i)
    {
        if((i==index || num[i]!=num[i-1])&&(num[i]<=left))
           {
               step[cnt]=num[i];
               dfs(i+1,cnt+1,left-num[i]);
           }
    }
    return;
}
int main()
{
    int tmp;
    while(scanf("%d%d",&t,&n)&&n)
    {
        for(int i=1;i<=n;++i)
        {
            scanf("%d",&tmp);
            if(tmp>t)
            {
                n--;
                i--;
            }
            else
            {
                num[i]=tmp;
            }
        }
        printf("Sums of %d:\n",t);
        No_Found=1;
        dfs(1,1,t);
        if(No_Found)
        {
            printf("NONE\n");
        }
    }
}

网上说这是一道水题，是吗？希望有一天我也这么说吧！

但我想这道题没想出来。