UVA 10125 Sumsets

Sumsets

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2549
64-bit integer IO format: %lld      Java class name: Main

 

iven 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 +536870911 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

 

解题：二分好啦

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
struct Node {
    LL val;
    int x,y;
    Node(LL v,int a,int b):val(v),x(a),y(b) {}
    bool operator<(const Node &t) const {
        return val < t.val;
    }
    bool operator>(const Node &t) const {
        return val > t.val;
    }
    bool operator!=(const Node &t) const {
        return x != t.x && y != t.y && x != t.y && y != t.x;
    }
};
vector<Node>a,b;
LL d[1010];
int main() {
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n && n) {
        for(int i = 0; i < n; ++i) cin>>d[i];
        a.clear();
        b.clear();
        for(int i = 0; i < n; ++i)
            for(int j = i + 1; j < n; ++j) {
                a.push_back(Node(d[i] + d[j],i,j));
                b.push_back(Node(d[i] - d[j],i,j));
                b.push_back(Node(d[j] - d[i],j,i));
            }
        sort(a.begin(),a.end());
        sort(b.begin(),b.end());
        LL ret;
        memset(&ret,0x80,sizeof(LL));
        for(auto it = a.begin(); it != a.end(); ++it){
            auto lvalue = lower_bound(b.begin(),b.end(),*it);
            auto rvalue = upper_bound(lvalue,b.end(),*it);
            for(;lvalue != rvalue; ++lvalue)
                if(*lvalue != *it)
                    ret = max(ret,it->val + d[lvalue->y]);
        }
        if (ret < -536870912)
            cout<<"no solution"<<endl;
        else cout<<ret<<endl;
    }
    return 0;
}
/*
5
2
3
5
7
12
*/