hdu 4611 Balls Rearrangement 数学

Balls Rearrangement

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

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

 

 

Author

SYSU

 

 

Source

2013 Multi-University Training Contest 2

题意：求sigma（i%a-i%b);   

思路：显然lcm为循环节；

　　   当同时模加1 的时候差值相等所以只有在对A或B取模为0的时候会改变；

　　　即A、B的倍数为断点；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=3e5+10,M=1e6+10,inf=1e9,mod=1e9+7;
ll a[N],b[N],c[N];
ll gcd(ll x,ll y)
{
    return y==0?x:gcd(y,x%y);
}
int main()
{
    ll x,y,z,i,t;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d",&x,&y,&z);
        ll gc=gcd(y,z);
        ll lcm=y*z/gc;
        int ji=1;
        for(i=1;i<=lcm/y;i++)
        a[ji++]=i*y;
        for(i=1;i<lcm/z;i++)
        a[ji++]=i*z;
        sort(a,a+ji);
        for(i=0;i<ji-1;i++)
        {
            b[i]=a[i+1]-a[i];
            c[i]=abs(a[i]%y-a[i]%z);
        }
        ll sum=0;
        for(i=0;i<ji-1;i++)
        sum+=b[i]*c[i];
        ll ans=0;
        ans+=x/lcm*sum;
        x%=lcm;
        for(i=0;i<ji-1;i++)
        {
            if(x<=0)
            break;
            if(x<b[i])
            ans+=x*c[i];
            else
            ans+=b[i]*c[i];
            x-=b[i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}