Problem DescriptionFat 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.
InputThe 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
OutputFor 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 Sample Output题意：
给出两个长整型的数a, b。有一种将a变化的操作为：a=a-(a%x), 其中 1<=x<=a-1;
问最少有多少次操作能使得a<=b。
#include <iostream>
#include <cstdio>
#include <cstring>
typedef long long ll;
using namespace std;
int main()
{
int t,count,k=1;
ll A,B;
cin>> t;
while(t--)
{
count=0;
cin>>A>>B;
while(A>B)
{
A=A/2;
count++;
}
printf("Case %d: %d\n",k++,count);
}
return 0;
}原文地址：http://blog.csdn.net/whjkm/article/details/45938335
