【哈希】1551:Sumsets

1551:Sumsets

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总时间限制: 

1000ms

内存限制: 

65536kB

描述

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.

输入

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.

输出

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

样例输入

5
2 
3 
5 
7 
12
5
2 
16 
64 
256 
1024
0

样例输出

12
no solution

来源

Waterloo local 2001.06.02

#include<stdio.h>
#include<queue>
#include<math.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int flag[1000010];//标记是否存有数据 
int data[1000010];//保存和 差
int h1[1000010];//加数1
int h2[1000010];//加数2
int in[1010];//输入数据
int a,b,c,d,n;
int hashh(int key){
	int ad=(key%1000000+1001234)%1000000;//取余函数
	while(flag[ad]==1){
		ad+=ad%11+1;
		if(ad>1000000)
			ad=ad%1000000;
	}
	return ad;
}
int find(int key){
	//返回key所在位置
	int ad=(key%1000000+1001234)%1000000;//取余函数
	while(flag[ad]==1&&data[ad]!=key){
		ad+=ad%11+1;
		if(ad>1000000)
			ad=ad%1000000;
	}
	return flag[ad]==0?-1:ad;
}
int main(){
	cin>>n;
	while(n!=0){
		memset(data,-1,sizeof(data));
		memset(flag,0,sizeof(flag));
		int ans=-536870912;
		for(int i=0;i<n;i++){
			cin>>in[i];
		}
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				//遍历a+b
				if(in[i]!=in[j]){
				int ad=hashh(in[i]+in[j]);
				flag[ad]=1;
				h1[ad]=in[i];
				h2[ad]=in[j];
				data[ad]=in[i]+in[j];
				}
			}
		}	
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				if(in[i]!=in[j]){
					int t=in[i]-in[j];//d-c
					int ad=find(t);
					if(ad!=-1&&in[i]!=h1[ad]&&in[i]!=h2[ad]&&in[j]!=h1[ad]&&in[j]!=h2[ad]){
						ans=max(ans,in[i]);
					}
				}
			}
		}
		if(ans!=-536870912)
			cout<<ans<<endl;
		else
			cout<<"no solution"<<endl;
		cin>>n;
	}
}