2015
09-17

# Balls Rearrangement

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.

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).

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).

3
1000000000 1 1
8 2 4
11 5 3

0
8
16

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
long long gcd(long long a,long long b){
return b ? gcd(b,a % b) : a;
}
long long lcm(long long a,long long b){
return a / gcd(a,b) * b;
}
long long get(long long n,long long a,long long b){
long long ans = 0,tem = 0,x = 0,y = 0,now = 0;
while(now < n){
tem = min(a - x,b - y);
if(tem + now > n)tem = n - now;
ans += tem * abs(x - y);
x = (x + tem) % a;
y = (y + tem) % b;
now += tem;
}
return ans;
}
int main(){
int t;
scanf("%d",&t);
while(t --){
long long n,a,b;
//scanf("%I64d%I64d%I64d",&n,&a,&b);
cin>>n>>a>>b;
long long ans = 0;
long long l = lcm(a,b);
if(n > l)ans = get(l,a,b) * (n / l) + get(n % l,a,b);
else ans = get(n,a,b);
//long long ans = get(lcm(a,b),a,b) * n / lcm(a,b) + get(n % lcm(a,b),a,b);
//printf("%I64d\n",ans);
cout<<ans<<endl;
}
return 0;
}