A-B Game_game a b c

Think:
这题 也是很醉啊！！！ 不知为何 用C的scanf就是WA 改成用C++的 Cin 就很神奇的AC了～～～～～～～～～ WA了14发2333333 然后 改着改着就神奇的对了～～～～～～～
Problem 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
2
5 3
10086 110
Sample Output
Case 1: 1
Case 2: 7

以下是 WA 代码 很神奇的 WA 不懂错在哪儿～～～～～～

#include<stdio.h>
int main()
{
    long long int a, b;
    int cnt,kase;
    int T;
    scanf("%d", &T);
        kase  =  0;
        while(T --)
        {
            scanf("%lld %lld", &a, &b);
            cnt = 0;
            kase = kase + 1;
            while(a > b)
            {
                a = a / 2 + 1;
                cnt ++;
            }
            printf("Case %d: %d\n", kase, cnt);
        }
}
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以下是 AC 代码 很神奇的改着改着 就AC了TAT

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    long long a, b;
    int kase = 0, cnt;
    int T;
    cin >> T;
    kase = 0;
    while(T --)
    {
        kase ++;
        cnt = 0;
        cin >> a >> b;
        while(a > b)
        {
            a = a / 2 + 1;
            cnt ++;
        }
        printf("Case %d: %d\n", kase, cnt);
    }
}
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