POJ 2549

Sumsets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8235		Accepted: 2260

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 +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

 

分成a + b  和 d - c两个集合

x = d - c    保存a + b 的所有值，二分查找x并判断合理性

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

struct node { int v,a,b; };
const int MAX = 1005;
int N,len = 0;
int ele[MAX];
node y[MAX * MAX];

bool cmp(node a,node b) { return a.v < b.v; }

bool solve() {
        for(int i = N - 1; i >= 0; --i) {
                for(int j = N - 1; j >= 0; --j) {
                        if(i == j) continue;
                        int d = ele[i] - ele[j];
                       // printf("ele = %d d = %d
",ele[i],d);
                        int l = 0,r = len - 1;
                        while(l < r) {
                                int mid = (l + r + 1) >> 1;
                                if(y[mid].v > d) r = mid - 1;
                                else l = mid;
                        }
                        if(y[l].v == d) {
                                for(int k = l; k < len; ++k) {
                                        if(y[k].v == d && y[k].a != ele[j] && y[k].b != ele[j]
                                           && y[k].a != ele[i] && y[k].b != ele[i]) {
                                                printf("%d
",ele[i]);
                                                return true;
                                        }
                                }
                                for(int k = l; k >= 0; --k) {
                                        if(y[k].v == d && y[k].a != ele[j] && y[k].b != ele[j]
                                           && y[k].a != ele[i] && y[k].b != ele[i]) {
                                                printf("%d
",ele[i]);
                                                return true;
                                        }

                                }

                        }

                }
        }

        return false;

}

int main()
{
   // freopen("sw.in","r",stdin);

    while(~scanf("%d",&N) && N ) {
            for(int i = 0; i < N; ++i) scanf("%d",&ele[i]);

            sort(ele,ele + N);
            N = unique(ele,ele + N) - ele;


            len = 0;
            for(int i = 0; i < N; ++i) {
                    for(int j = i + 1; j < N; ++j) {
                            y[len].v = ele[i] + ele[j];
                            y[len].a = ele[i];
                            y[len++].b = ele[j];
                    }
            }

            sort(y,y + len,cmp);

            if(!solve()) printf("no solution
");
    }
    return 0;
}