A-B Game FZU - 2147

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本文介绍了一个名为A-BGameFZU-2147的特殊游戏算法问题，该问题涉及两个整数A和B。文章详细解释了游戏规则及如何通过改变A的值来结束游戏，并给出了一个有效的算法实现，通过选择特定的除数以减少操作次数。

A-B Game

FZU - 2147

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

2
5 3
10086 110

Sample Output

Case 1: 1
Case 2: 7

每次模a/2+1是最大的值

不能用%lld wa了三次，用cin 读入或者%I64读入就对了

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int t,cas = 0;
    __int64 a,b;
    scanf("%d",&t);
    while(t--){
        scanf("%I64d%I64d",&a,&b);
        __int64 tmp;
        int cnt = 0;
        while(a > b){
            tmp = a / 2 + 1;
            a = a - (a % tmp);
            cnt++;
        }
        printf("Case %d: %d\n",++cas,cnt);
    }
    return 0;
}