hdu4611 Balls Rearrangement

Balls Rearrangement

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1682 Accepted Submission(s): 634

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 Multi-University Training Contest 2

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zhuyuanchen520

唉，比赛的时候，一直想通过推出公式来，纠结在起过一个周期时，找到规律，其实，我们只要把有统一差值的一起算，不是统一差值的下次再算就已经很快了啊！

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
__int64 gcd(__int64 a,__int64 b)
{
    if(a==0)return b;
    return gcd(b%a,a);
}
__int64 solve(__int64 n,__int64 a,__int64 b)
{
    __int64 sum=0,k1=0,k2=0,i=0,temp;
    while(i<n)
    {
        temp=min(a-k1,b-k2);
        i+=temp,sum+=temp*fabs(k1-k2);
        if(i>n)
        sum-=(i-n)*fabs(k1-k2);
        k1=(k1+temp)%a,k2=(k2+temp)%b;
    }
    return sum;
}
int main()
{
    int tcase;
    __int64 n,m,a,b;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%I64d%I64d%I64d",&n,&a,&b);
        m=a*b/gcd(a,b);
        if(n<m)
        {
            printf("%I64d\n",solve(n,a,b));
        }
        else
            printf("%I64d\n",n/m*solve(m,a,b)+solve(n%m,a,b));
    }
    return 0;
}