题目来源：http://acm.fzu.edu.cn/contest/problem.php?cid=134&sortid=8
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： a= A - (A %x) ,我们知道 a 是递减函数， 直到=<b 。因此我们需要 求 a的最小值 ， 即求 （A%x）的最大值， 经过测试，发现 （A%x） 的最大值 是 （A-1）/2。
代码如下：
#include<iostream> #include<stdlib.h> #include<stdio.h> #include<math.h> #include<string.h> #include<string> #include<queue> #include<algorithm> #include<map> using namespace std; typedef long long LL; int solve(LL a, LL b) { int num=0; while(a>b) { num++; a=a-(a-1)/2; } return num; } int main() { LL m,n; int t,k=1; cin>>t; while(t--) { cin>>m>>n; printf("Case %d: ",k++); cout<<solve(m,n)<<endl; } return 0; }
A-B Game 水题+longlong,布布扣,bubuko.com
原文：http://www.cnblogs.com/zn505119020/p/3601042.html