FZU A-B 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

25 310086 110

Sample Output

Case 1: 1Case 2: 7

题目分析：

A -  (A%x)  ,x 取过大或者过小都不合适，所以最好的是取中间

#include <iostream>#include <cstdio>#include <cstring>#include <iomanip>using namespace std;long long  a, b;long t;void solve(long long a, long long  b){    long sum = 0;    while (a > b)    {        a -= ((a - 1) / 2);        sum++;    }    printf("%d\n", sum);}int main(){    cin>>t;    for (long i = 1; i <= t; i++)    {        scanf("%I64d%I64d\n", &a, &b);        printf("Case %d: ", i);        solve(a, b);    }    return 0;}