hdu 4611 Balls Rearrangement （找循环节）

Balls Rearrangement

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

Total Submission(s): 1700    Accepted Submission(s): 643

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

31000000000 1 18 2 411 5 3

 

Sample Output

0816

 

Source

2013 Multi-University Training Contest 2

 

思路：

将0~N-1分别写成两个矩阵的形式其实就有思路了，每次充满两矩阵中的其中一列，因为初始间隔一定，每走一步或每走多步的相对间隔仍然没变，所以可以一次统计出来（即每次用i+=step代替i++）。这样还有一种情况不能处理，就是两个矩形的宽都很小（即n很大，a、b很小），所以想到找循环节，循环节很容易找即lcm(a,b)。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;ll n,m,a,b,ans,lcm;ll gcd(ll x,ll y){    ll r=x%y;    if(r==0) return y;    return gcd(y,r);}void solve(){    ll i,j,r,mod,sum,t;    ll p1,p2,step,mi;    r=n/lcm;    mod=n%lcm;    p1=p2=0;    sum=t=0;    for(i=0;i<lcm;i+=step)    {        step=min(a-p1,b-p2);      // 充满矩形的一列需要走的步数        if(i<mod&&i+step+1>=mod)        {            t=sum+(mod-i)*fabs(p1*1.0-p2);        }        sum+=step*fabs(p1*1.0-p2);        p1=(p1+step)%a;           // 更新位置指针        p2=(p2+step)%b;    }    ans=r*sum+t;}int main(){    ll i,j,t,c;    scanf("%I64d",&t);    while(t--)    {        scanf("%I64d%I64d%I64d",&n,&a,&b);        c=gcd(a,b);        lcm=a/c*b;        solve();        printf("%I64d\n",ans);    }    return 0;}