首页 > ACM题库 > HDU-杭电 > HDU 1258 Sum It Up-DFS-[解题报告] c-sharp
2013
12-04

HDU 1258 Sum It Up-DFS-[解题报告] c-sharp

Sum It Up

问题描述 :

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

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1258
题目描述:给你一个数t作为最后等式的和,并给你一组数a[i](i<12)作为等式的加数,每个加数最多只能使用一次,要求输出所有满足条件(加数从大到小输出)的等式,并且不能重复。
例:t=4. a[]={4,3,2,2,1,1}
输出 4,3+1,2+2,2+1+1
解题思路:题目要求输出从大到小输出,所以可以先给所有的数进行降序排序。数据的规模很小,所以可以用暴力搜索。搜索过程中,为了避免重复输出,需要记录前一层搜索的起点,下一层递归搜索的起点不能与前一层记录的点一样…说不清楚,具体见代码

 

#include <iostream>
#include <algorithm>
using namespace std;

int t,n;
int a[20];
int save[20]; 
int index;
int used[20];
int sign;

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

void dfs(int k, int sum)
{
	if(sum > t)
	{
		return ;
	}
	if(sum == t)
	{
		sign = 1;
		for(int i=0; i<index-1; i++)
		{
			cout<<save[i]<<"+";
		}
		cout<<save[index-1]<<endl;
		return ;
	}
	int last = -1;
	for(int i=k+1; i<=n; i++)
	{
		if(a[i] != last)           //当前的数不能跟上一次搜索的起点的数值一样,不然会造成重复
		{
			save[index++] = a[i];
			last = a[i];            //last保存当前搜索的起点
			dfs(i,sum+a[i]);
			index--;
		}
	}
}

int main()
{
	int i;
	while(cin>>t>>n,t+n)
	{
		index = 0;
		sign = 0;
		for(i=1; i<=n; i++)
		{
			cin>>a[i];
		}
		sort(a+1,a+n+1,cmp); //降序排序
		printf("Sums of %d:/n",t);
		dfs(0,0);
		if(!sign)
		{
			cout<<"NONE"<<endl;
		}
	}
	return 0;
}

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