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
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 "); 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 ",ans); 31 } 32 } 33 return 0; 34 }