Greatest Number

时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte

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<M≤ 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.

 

样例输入

2 10

100 2

0 0

 

样例输出

Case 1: 8

题意：给出n个数，每个数可互相叠加，最多叠加四次，问最后叠加的数的最大值是多少，ps:最大值不能超过m。

题解：先枚举每两个数叠加的值(不超过m)，用数组str[]保存，题目可简化为叠加一次str，这样即可看做叠加了四次,

     但在叠加过程中需要查找叠加后的和不超过m的数，故用二分可不超时.

 

 

#include
        
         #include
         
          #include
          
           #include
           
            #include
            
             using namespace std;int str[1000010],op[1010];int main(){    int i,j,k,n,m,ans,t=1;    bool flag=false;    while(~scanf("%d%d",&n,&m),n||m)    {        op[0] = k = 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);        ans = 0;        for(i=0;i
             
              m) high = mid-1; else low = mid+1; } if(str[i]+str[low]>ans && str[i]+str[low]<=m) ans = str[i]+str[low]; } if(flag) printf("\n"); printf("Case %d: %d\n",t++,ans); flag = true; } return 0;}