HDU 4710 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.

 

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

31000000000 1 18 2 411 5 3 

 

Sample Output

0816 

打表可知 求出a b的最小公倍数 为一个周期

i%a-i%b 如果分成一段一段的话 跟i的大小可以无关  下面用的是tmp记录 然后now更新走了多长 x,y也不段更新 就可求出答案

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