FZU-2147-2147 A-B Game,规律题。。

Problem 2147 A-B Game

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

    

   规律很好找，求多少次A-(A%X)<=B;也就是求A%X的最大值，这样减的次数才会尽可能少，于是枚举了几组数据，发现当X=A/2+1时，A%X最接近A;于是规律出来了；

      

#include
#include
#include
#include
#include
using namespace std;
//const int N=101;
//char a[N][N];
int main()
{
   int t,i;
   long long a,b,x;
   scanf("%d",&t);
   int t1=t;
   while(t--)
   {
       x=i=0;
       scanf("%I64d%I64d",&a,&b);
         while(a>b)
       {
           i++;
           x=a/2+1;//关键找到这个规律；
           a-=a%x;
       }
       printf("Case %d: ",t1-t);
       printf("%d\n",i);
   }
    return 0;
}

转载于:https://www.cnblogs.com/nyist-TC-LYQ/p/7208295.html