http://poj.org/problem?id=2549

维基百科有3Sum算法：https://en.wikipedia.org/wiki/3SUM

 sort(S);
 for i=0 to n-3 do
    a = S[i];
    k = i+1;
    l = n-1;
    while (k<l) do
       b = S[k];
       c = S[l];
       if (a+b+c == 0) then
          output a, b, c;
          // Continue search for all triplet combinations summing to zero.
           k = k + 1
           l = l - 1
       else if (a+b+c > 0) then
          l = l - 1;
       else
          k = k + 1;
       end   
    end
 end

1、排序，求任意两个数的和；将和与（i,j）通过Hash数组做映射，处理hash冲突用图论中的邻接表---学数据结构的时候老师讲过的桶式哈希

2、枚举两数(p,q)之差，通过HASH找到（i,j），保证p,q,i,j都不等即可

40多ms

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdin)

const int SIZE = 536870911;
const int MOD = 9973;
const int MAXN = 1000 + 100;

ll s[MAXN];
struct Node
{
    int i;
    int j;
    int next;
}rec[1100005];
int Hash[1100005];//头结点
inline ll Abs(ll x)
{
    return x>0?x:(-x);
}

int main()
{
    //IN("poj2549.txt");
    int n,cnt;
    while(~scanf("%d",&n) && n)
    {
        CL(Hash,0xff);
        for(int i=0;i<n;i++)
            scanf("%lld",&s[i]);
        sort(s,s+n);
        cnt=0;
        for(int i=0;i<n-1;i++)
            for(int j=i+1;j<n;j++)
            {
                rec[cnt].i=i;
                rec[cnt].j=j;
                ll sum=Abs(s[i]+s[j])%MOD;
                rec[cnt].next=Hash[sum];
                Hash[sum]=cnt;
                cnt++;
            }
        int ok=0;
        ll ans=0;
        for(int i=n-1;i>0;i--)
        {
            if(ok)break;
             for(int j=0;j<i;j++)
            {
                if(ok)break;
                ll sum=Abs(s[i]-s[j])%MOD;
                for(int v=Hash[sum];v!=-1;v=rec[v].next)
                {
                    int h=rec[v].i;
                    int k=rec[v].j;
                    if(s[h]+s[k]==s[i]-s[j] && rec[v].i!=i && rec[v].i!=j && rec[v].j!=j && rec[v].j!=i)
                    {
                        ok=1;
                        ans=s[i];
                        break;
                    }
                }
            }
        }
        if(ok)printf("%lld\n",ans);
        else printf("no solution\n");
    }
    return 0;
}