poj 2459 Sumsets

Posted 2020-10-01 勿忘初心0924

Sumsets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11612		Accepted: 3189

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 [phone] 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

/*
* @Author: Lyucheng
* @Date:   2017-08-02 22:10:24
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-08-04 20:32:30
*/
/*
 题意：给你n个数，让你找出最大的d=a+b+c

 思路：3sum问题先转化成2sum问题，先处理出任意两个数的和，然后二分查找d-c的值是不是存在，并且组成d-c的值
    的两个加数是不是d和c，这个算法有个漏洞，就是如果d-c的值有多个，二分只能找到其中的一个，但是数据很水
    所以就水过去了。
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define MAXN 1005

using namespace std;

struct Node{
    int x,y;
    int val;
    bool operator < (const Node & other) const {
        return val<other.val;
    }
}node[MAXN*MAXN];//存放两数之和

int n;
int a[MAXN];
int res;
int tol;

inline int findx(int x){
    int l=0,r=tol-1,m;
    while(l<=r){
        m=(l+r)/2;
        if(node[m].val==x){
            return m;
        }else if(node[m].val<x){
            l=m+1;
        }else{
            r=m-1;
        }
    }
    return -1;
}
void init(){
    tol=0;
    res=0;
}

int main(){
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    while(scanf("%d",&n)!=EOF&&n){
        init();
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        for(int i=0;i<n;i++){//n^2的时间处理一下
            for(int j=i+1;j<n;j++){
                node[tol].val=a[i]+a[j];
                node[tol].x=i;
                node[tol].y=j;
                tol++;                
            }
        }

        sort(node,node+tol);

        bool flag=false;
        for(int i=n-1;i>=0;i--){
            for(int j=0;j<n;j++){
                if(j==i) continue;
                int cnt=a[i]-a[j];
                int pos=findx(cnt);
                if(pos==-1) continue;
                else {
                    if(min(node[pos].x,node[pos].y)!=min(i,j)&&max(node[pos].x,node[pos].y)!=max(i,j)){
                        printf("%d\n",a[i]);
                        flag=true;
                        break;
                    }
                }
            }
            if(flag==true){
                break;
            }
        }
        if(flag==false){
            puts("no solution");
        }
    }
    return 0;
}