POJ

条件a+b == d-c，O(n*n)枚举出所有的形如a+b的和，然后从大到小枚举d，在枚举c，二分找d-c。

主要是要用容斥排除c和d出现在和中，详细的过程见注释。

复杂度为O(n^2log(n^2))。

/*********************************************************
*            ------------------                          *
*   author AbyssalFish                                   *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;

/*
容斥原理去重复
左边是枚举出来的,保证不等
可能出现重复的情况有：
d 和 c同时出现
case 1 d+c == d-c -> c == 0

其中一个出现
前提
a or b != c or d

case2  a+d == d-c, -> a == -c , 必要条件 a != c -> c != 0 如果不满足这个条件：找到的可能是d == d
case3  d+b == d-c, -> b == -c
case 2 3互斥

case4  a+c == d-c, -> a == d-2*c, 必要条件 a != d -> c != 0
case5  c+b == d-c -> b == d-2*c
case 4 5互斥

case (2,3,4,5) 1互斥
*/

const int maxn = 1e3, INF = 536870911+5;
int n;
int e[maxn+1], tab[maxn*maxn>>1];
bool vis[maxn];

int solve()
{
    sort(e,e+n);
    int i, j, a, c, d, k;
    int sz = 0;
    for(i = 1; i < n; i++){
        a = e[i];
        for(j = 0; j < i; j++){
            tab[sz++] = a+e[j];
        }
    }
    sort(tab,tab+sz);
    e[n] = e[n]-1;
    for(j = 0; j < n; j++) {
        c = -e[j];
        vis[j] = (*lower_bound(e,e+n,c) == c); // c == 0 || -c == a or b
    }
    tab[sz+1] = tab[sz] = tab[sz-1]-1;
    for(i = n; i--; ){
        d = e[i];
        for(j = 2; j < n; j++)if(i!=j){
            c = e[j];
            a = d-c;
            k = lower_bound(tab,tab+sz,a)-tab;
            if(c){
                if(vis[j]) k++;
                if(*lower_bound(e,e+n,a-c) == a-c) k++;
            }
            else k++;
            if(a == tab[k]) return d;
        }
    }
    return INF;
}

//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    while(scanf("%d",&n),n){
        for(int i = 0; i < n; i++) scanf("%d", e+i);
        int res = solve();
        if(res < INF) printf("%d
", res);
        else puts("no solution");
    }
    return 0;
}

 伪二分O(n^3) ???

int solve()
{
    sort(e,e+n);
    int i, j, sum, lb, ub, tsum;

    for(i = n; i--; ){
        for(j = 2; j < n; j++)if(i!=j){
            sum = e[i]-e[j];
            for(lb = 0,ub = j-1; lb < ub; ){ //状态是DAG
                tsum = e[lb]+e[ub];
                if(tsum == sum) return e[i];
                tsum > sum ? ub--:lb++; //O(n)
            }
        }
    }
    return INF;
}