hduoj 4710 Balls Rearrangement 2013 ACM/ICPC Asia Regional Online —— Warmup

http://acm.hdu.edu.cn/showproblem.php?pid=4710

 

Balls Rearrangement

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 735    Accepted Submission(s): 305

Problem Description

Bob has N balls and A boxes. He numbers the balls from 0 to N-1, and numbers the boxes from 0 to A-1. To find the balls easily, he puts the ball numbered x into the box numbered a if x = a mod A. Some day Bob buys B new boxes, and he wants to rearrange the balls from the old boxes to the new boxes. The new boxes are numbered from 0 to B-1. After the rearrangement, the ball numbered x should be in the box number b if x = b mod B. This work may be very boring, so he wants to know the cost before the rearrangement. If he moves a ball from the old box numbered a to the new box numbered b, the cost he considered would be |a-b|. The total cost is the sum of the cost to move every ball, and it is what Bob is interested in now.

 

 

Input

The first line of the input is an integer T, the number of test cases.(0<T<=50) Then T test case followed. The only line of each test case are three integers N, A and B.(1<=N<=1000000000, 1<=A,B<=100000).

 

 

Output

For each test case, output the total cost.

 

 

Sample Input

3 1000000000 1 1 8 2 4 11 5 3

 

 

Sample Output

0 8 16

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

 

分析：

 

模拟，每次增加 step ，一次可以放一块。

 

AC代码：

 

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<math.h>

 4 #define min(a,b) a>b?b:a

 5 using namespace std;

 6 int main()

 7 {

 8     int T,n,a,b;

 9     cin>>T;

10     while(T--)

11     {

12         cin>>n>>a>>b;

13         if(a==b)

14             printf("0\n");

15         else

16         {

17             __int64 ans=0,step=1,i;

18             for(i=0;i<n;i=i+step)

19             {

20                 int stepa=a-i%a;

21                 int stepb=b-i%b;

22                 step=min(stepa,stepb);

23                 __int64 dis=abs(i%a-i%b);

24                 if(i+step>=n)

25                     dis=dis*(n-i);

26                 else

27                     dis=dis*step;

28                 ans=ans+dis;

29             }

30             printf("%I64d\n",ans);

31         }

32     }

33     return 0;

34 }

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