2015
09-17

# Balls Rearrangement

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 344    Accepted Submission(s): 165

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：对于i%a - i%b,每次加上从i开始和这个值(i%a - i%b)相等的一段,

*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<cmath>
#include<math.h>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=10;
__int64 p;

__int64 Gcd(__int64 a,__int64 b){
if(b == 0)return a;
return Gcd(b,a%b);
}

__int64 calculate(__int64 n,__int64 a,__int64 b,__int64 num){
p=0;
__int64 la=a,lb=b,sum=0,l;
for(__int64 i=0;i<n;){
l=min(la,min(lb,n-i));
if(i+l>num && i<num)p=sum+abs((int)(i%a - i%b))*(num-i);
sum+=abs((int)(i%a - i%b))*l;
i+=l;
la=(la-l+a-1)%a+1;
lb=(lb-l+b-1)%b+1;
}
return sum;
}

int main(){
__int64 n,a,b,t;
scanf("%I64d",&t);
while(t--){
scanf("%I64d%I64d%I64d",&n,&a,&b);
__int64 gcd=Gcd(a,b),per=a*b/gcd,k=min(per,n);//求出最小公倍数
__int64 sum=calculate(k,a,b,n%k);
if(n>per)sum=(n/per)*sum+p;//p表示前n%k个i%a-i%b的和
printf("%I64d\n",sum);
}
return 0;
}