poj 2549(二分)

这题的关键在于将  a+b+c=d 转换为  a+b=d-c  ， 然后将所有a+b 的情况全部求出来排序， 然后从大到小枚举d和c， 二分查找d-c。

 

Sumsets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7682		Accepted: 2105

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

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;

struct node
{
    int x,y,key;
}k[1001000];

int g[1010];
int s;

int cmp(node t,node t1)
{
    return t.key<t1.key;
}

int pan(int i,int j,int mid)
{
    if(k[mid].x!=g[i]&&k[mid].y!=g[i]&&k[mid].x!=g[j]&&k[mid].y!=g[j])
    {
        return 1;
    }
    return 0;
}

int main()
{
    while(scanf("%d",&s)&&s)
    {
        for(int i=0;i<s;i++)
            scanf("%d",&g[i]);
        int cnt=0;
        for(int i=0;i<s-1;i++)
            for(int j=i+1;j<s;j++)
            {
                k[cnt].x=g[i]; k[cnt].y=g[j];
                k[cnt].key=g[i]+g[j];
                cnt++;
            }
        sort(k,k+cnt,cmp);
        sort(g,g+s);
        int flag=0,tmp,b,d,mid;
        for(int i=s-1;i>=0;i--)
        {
            for(int j=s-1;j>=0;j--)
            {
                if(i==j) continue;
                tmp=g[i]-g[j];
                b=0; d=cnt-1;
                while(b<=d)
                {
                    mid=(b+d)/2;
                    if(k[mid].key==tmp)
                    {
                        for(int i1=max(0,mid-1);i1<=min(mid+1,cnt-1);i1++)
                        {
                            if(k[i1].key==tmp && pan(i,j,i1)==1)
                            {
                                flag=1;
                                break;
                            }
                        }
                        break;
                    }
                    if(k[mid].key>tmp)
                        d=mid-1;
                    else b=mid+1;
                }
                if(flag==1)
                {
                    printf("%d\n",g[i]);
                    break;
                }
            }
            if(flag==1) break;
        }
        if(flag==0) printf("no solution\n");
    }
    return 0;
}