ACM--贪心--数学--A-B Game--水

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OJ地址：http://acm.fzu.edu.cn/problem.php?pid=2147

Problem 2147 A-B Game

Accept: 1158 Submit: 2769
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

25 310086 110

Sample Output

Case 1: 1Case 2: 7

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题意：给出两个数A,B（A>B），每次操作A-(A%x),x任选，问至少需要几次操作能够使得A<B。

解法：简单贪心。每次x都取A/2+1。

代码:

 #include<iostream> #include<cstdio> #include<stdio.h> using namespace std; /**    输入A、B（B<A）。A每次可以进行一次变换：A = A - (A % x)    ，其中1≤x≤A-1。问最少经过多少次变换，A的值小于等于B。    这题的要点就是x的取值上面，每次去x=A/2+1是最好的 */ int main(void){    int n,index=0;    long long  A,B;    scanf("%d",&n);    while(n--){      int result=0;      cin>>A;      cin>>B;      index++;      while(A>B){         A=A-(A%(A/2+1));         result++;      }      printf("Case %d: ",index);      cout<<result<<endl;    }     return 0; }

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