1341-A-B Game(找规律) ZCMU

最新推荐文章于 2022-08-05 18:07:43 发布

原创最新推荐文章于 2022-08-05 18:07:43 发布 · 315 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#规律题

练习专栏收录该内容

160 篇文章

订阅专栏

本文介绍了一个特殊的二人游戏算法问题：给定两个整数A和B，玩家通过一系列操作使A减少到小于或等于B，目标是最小化操作次数。文章提供了输入输出样例及解析程序代码。

Description

Fat brother and Maze are playing a kind of special (hentai) game by two integers A and B. First Fat brother write an integer A on a white paper and then Maze start to change this integer. Every time Maze can select an integer x between 1 and A-1 then change A into A-(A%x). The game ends when this integer is less than or equals to B. Here is the problem, at least how many times Maze needs to perform to end this special (hentai) game.

Input

The first line of the date is an integer T, which is the number of the text cases.

Then T cases follow, each case contains two integers A and B described above.

1 <= T <=100, 2 <= B < A < 100861008610086

Output

For each case, output the case number first, and then output an integer describes the number of times Maze needs to perform. See the sample input and output for more details.

Sample Input

5 3

10086 110

Sample Output

Case 1: 1

Case 2: 7

代码

#include<stdio.h>
int main()
{
    int T,kase=1;
    scanf("%d",&T);
    while(T--)
    {
        long long a,b,cnt;
        scanf("%lld%lld",&a,&b);
        cnt=0;
        while(a>b)
        {
            if(a%2==0)
                a-=a/2-1;
            else
                a-=a/2;
            cnt++;
        }
        printf("Case %d: %lld\n",kase++,cnt);
    }
    return 0;
}