poj--2549--Sumsets(二分查找)
| Time Limit: 1000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u | 
Description
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 5 7 12 5 2 16 64 256 1024 0
Sample Output
12 no solution
Source
Waterloo local 2001.06.02
/*在数列中找四个数使a+b+c=d;转化之后得a+b=d-c,先对数列排序列举每一个d-c,
二分查找a和b*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX 100100
#define INF -0x3f3f3f
int a[MAX];
int main()
{
	int n;
	while(scanf("%d",&n),n)
	{
		for(int i=0;i<n;i++)
		scanf("%d",&a[i]);
		sort(a,a+n);
		int ans=INF;
		for(int i=n-1;i>=0;i--)
		{
			for(int j=n-1;j>=0;j--)
			{
				if(i==j)
				continue;
				int sum=a[i]-a[j];
				int l=0,r=j-1;
				while(l<r)
				{
					if(a[l]+a[r]==sum)
					{
						ans=a[i];
						break;
					}
					 if(a[l]+a[r]>sum)
					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;
} 
                    
                
 
 
                
            
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浙公网安备 33010602011771号