山东省第一届ACM大学生程序设计竞赛--Greatest Number--二分

Greatest Number

Time Limit: 1000MS Memory limit: 65536K

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 10
100
2

0 0

示例输出

Case 1: 8

提示

 给出n个数，从中找到4个或4个以内的数相加的和接近m（可以重复）

定义两个相同的数组，第一部分 a[0] = 0 ; 第二部分n个数，第三部分n个数任意两个相加的和，然后遍历其中一个，用二分的办法从第二个中找出和最接近m的 （保留和），最后得到最大的和

来源

 2010年山东省第一届ACM大学生程序设计竞赛

示例程序

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[2000001];
int cmp(int x, int y)
{
    return x < y;
}
int main()
{
    int n, m, i, j, mid, l, h, k, max1, x, num=0, g, r, z;
    while(scanf("%d%d",&n,&m)!=EOF&&n)
    {
        k=0;
        num++;
        a[0]=0;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            {
                if(a[i]>m)
                {
                    i--;
                    n--;
                }
            }
        }
        k=n+1;
        for(i=1; i<=n; i++)
        {
            for(j=i; j<=n; j++)
            {
                if(a[i]+a[j]<=m)
                {
                    a[k++]=a[i]+a[j];
                }
            }
        }
        sort(a,a+k,cmp);
        r=k-1;
        g=k-1;
        max1=0;
        for(i=0; i<k; i++)
        {
            z=0;
            l=i;
            while(l<=r)
            {
                mid=(l+r)/2;
                x=a[i]+a[mid];
                if(x>m)
                    r=mid-1;
                else if(x==m)
                {
                    max1=m;
                    z=1;
                    break;
                }
                else
                {
                    l=mid+1;
                    if(max1<x)
                    {
                        g=mid;
                        max1=x;
                    }
                }
            }
            if(z)
                break;
            r=g;
        }
        printf("Case %d: %d\n\n",num, max1);
    }
    return 0;
}