A-B Game 水题+longlong

题目来源：http://acm.fzu.edu.cn/contest/problem.php?cid=134&sortid=8

Problem H A-B Game

Accept: 103    Submit: 237
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 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

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;
}