山东省第一届ACM省赛题——Greatest Number（不用dp的二分查找）

题目描述

Saya likes math, because she think math can make her cleverer.
One day, Kudo invited a very simple game:
Given N integers, then the players choose no more than four integers from them (can be repeated) and add them together. Finally, the one whose sum is the largest wins the game. It seems very simple, but there is one more condition: the sum shouldn’t larger than a number M.
Saya is very interest in this game. She says that since the number of integers is finite, we can enumerate all the selecting and find the largest sum. Saya calls the largest sum Greatest Number (GN). After reflecting for a while, Saya declares that she found the GN and shows her answer.
Kudo wants to know whether Saya’s answer is the best, so she comes to you for help.
Can you help her to compute the GN?

输入

The input consists of several test cases.
The first line of input in each test case contains two integers N (0<N≤1000) and M(0 1000000000), which represent the number of integers and the upper bound.
Each of the next N lines contains the integers. (Not larger than 1000000000)
The last case is followed by a line containing two zeros.

输出

For each case, print the case number (1, 2 …) and the GN.
Your output format should imitate the sample output. Print a blank line after each test case.

示例输入

2 1010020 0

示例输出

Case 1: 8

提示

 

来源

 2010年山东省第一届ACM大学生程序设计竞赛

最多选四个数来相加不大于m，一开始我就想出dp，但数据太大，看了看大神的才知道用二分查找来加速，算法非常巧妙，值得回味

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>using namespace std;int str[10000000],op[1005];int main(){    int n,m,i,j,k,mid,low,high,t=1;    while(~scanf("%d%d",&n,&m)&&(n&&m))    {        k=0;        op[0]=0;        for(i=1; i<=n; ++i)            scanf("%d",&op[i]);        sort(op,op+n+1);        for(i=0; i<=n; ++i)            for(j=i; j<=n; ++j)            {                if(op[i]+op[j]<m)                    str[k++]=op[i]+op[j];            }        sort(str,str+k);        int ans=0;        for(i=0; i<k; ++i)        {            low=i,high=k-1;            while(low<high)            {                mid=(low+high)/2;                if(str[i]+str[mid]>m)                    high=mid-1;                else if(str[i]+str[mid]<m)                    low=mid+1;                else                    break;            }            if(str[i]+str[high]>ans&&str[i]+str[high]<=m)  //唯独不能用mid                ans=str[i]+str[high];        }        printf("Case %d: %d\n\n",t++,ans);    }    return 0;}