Greatest Number

最新推荐文章于 2024-04-22 14:04:25 发布

原创最新推荐文章于 2024-04-22 14:04:25 发布 · 545 阅读

1 ·

CC 4.0 BY-SA版权

练习专栏收录该内容

389 篇文章

订阅专栏

本文探讨了在游戏开发中引入数学策略的概念，通过一个简单的游戏实例，展示了如何利用有限的整数集合，通过不超过四次的加法操作，寻找最大可能的和值。文章详细解释了算法实现过程，并提供了示例代码，帮助开发者理解和实现这一有趣的游戏逻辑。

Greatest Number

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

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

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

示例程序

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<algorithm>  
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<k;i++)  
        {  
            int low=i,high=k-1;  
            while(low<high)  
            {  
                int mid=(low+high)/2;  
                if(str[i]+str[mid]>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;  
}