poj2549

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
#include<set>
#include<bitset>
//#define ONLINE_JUDGE
#define eps 1e-6
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
#define ll long long
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
//#pragma comment(linker,"/STACK:1024000000,1024000000")
int n,m;
#define N 210
#define M 1000010
#define Mod 1000000000
#define p(x,y) make_pair(x,y)
#define pr pair<ll,int>
struct Node{
int sum;
int id1,id2;
bool operator < (const Node &x) const{
return sum<x.sum;
}
};
vector<Node> v1;
vector<Node> v2;
int a[1010];
bool check(int id1,int id2,int id3,int id4){
if(id1 == id2 || id1 == id3 || id1 == id4)
return false;
if(id2 == id1 || id2 == id3 || id2 == id4)
return false;
if(id3 == id1 || id3 == id2 || id3 == id4)
return false;
if(id4 == id1 || id4 == id2 || id4 == id3)
return false;
return true;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
#endif
while(sf(n)!=EOF&&n){
for(int i=0;i<n;i++)
sf(a[i]);
Node tmp;
v1.clear();
v2.clear();
for(int i=0;i<n;i++){       //循环确定所有a+b的值
for(int j=i+1;j<n;j++){
tmp.sum = a[i]+a[j];
tmp.id1 = i;
tmp.id2 = j;
v1.push_back(tmp);
}
}
for(int i=0;i<n;i++){       //循环确定所有d-c值
for(int j=0;j<n;j++){
if(i == j) continue;
tmp.sum = a[i]-a[j];
tmp.id1 = i;
tmp.id2 = j;
v2.push_back(tmp);
}
}
sort(v1.begin(),v1.end());
sort(v2.begin(),v2.end());  //排序
int size = (int)v1.size();
int size2 = (int)v2.size();
int ans=-INF;
for(int i=0;i<size;i++){
int tmp = v1[i].sum;
int id1 = v1[i].id1;
int id2 = v1[i].id2;

int id = lower_bound(v2.begin(),v2.end(),v1[i])-v2.begin();//二分寻找第一个>=a+b的值

while(id<size2 && v2[id].sum == tmp){ //从第一个>=a+b的值到第一个>a+b的值位置，其中所有都符合a+b+c=d
int id3 = v2[id].id1;
int id4 = v2[id].id2;
if(check(id1,id2,id3,id4)){ //a,b,c,d四个数的id都不相同
ans = Max(ans,a[id3]);
}
id++;
}
}
if(ans != -INF)
printf("%d\n",ans);
else
printf("no solution\n");
}
return 0;
}