POJ 2549 Sumsets(折半枚举)

Sumsets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9892		Accepted: 2705

Description

Given S, a set of integers, find the largest d such that a + b + c = d where a, b, c, and d are distinct elements of S.

Input

Several S, each consisting of a line containing an integer 1 <= n <= 1000 indicating the number of elements in S, followed by the elements of S, one per line. Each element of S is a distinct integer between -536870912 and +536870911 inclusive. The last line of input contains 0.

Output

For each S, a single line containing d, or a single line containing "no solution".

Sample Input

  
  
  
  

   
   
   
   5
  
  
  
  
  
  
  
  

   
   
   
   2 
  
  
  
  
3 

  
  
  
  

   
   
   
   7 
  
  
  
  
5 
12

  
  
  
  

   
   
   
   16 
  
  
  
  
5
2 

  
  
  
  

   
   
   
   256 
  
  
  
  
64 
1024

  
  
  
  

   
   
   
   0
  
  
  
  

Sample Output

  
  
  
  

   
   
   
   12
  
  
  
  
  
  
  
  

   
   
   
   no solution
  
  
  
  

题意：在集合S中有n个数，找到最大的d,且d满足于集合内a+b+c=d。

题解：我们把找a+b+c=d化为找 a+b=d-c。设c为a，b，c中最大的元素。  注意d不一定比c大，d-c可以为负数。 这样我们枚举d，c，利用二分的思想查找a，b。 

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f 
using namespace std;
int a[1010];
int main()
{
	int n,i,j;
	while(scanf("%d",&n)&&n)
	{
		for(i=0;i<n;++i)
			scanf("%d",&a[i]);
		sort(a,a+n);
		int ans=INF,cnt;
		for(i=n-1;i>=0;--i)
		{
			for(j=n-1;j>=0;--j)
			{
				if(i==j)
					continue;
				cnt=a[i]-a[j];
				for(int l=0,r=j-1;l<r;)
				{
					if(a[l]+a[r]==cnt)
					{
						ans=a[i];
						break;
					}
					if(a[l]+a[r]>cnt)
						r--;
					else
						l++;
				}
				if(ans!=INF)
					break; 
			}
			if(ans!=INF)
				break;
		}
		if(ans!=INF)
			printf("%d\n",ans);
		else
			printf("no solution\n");
	}
	return 0;
}