2013 多校联合 2 A Balls Rearrangement (hdu 4611)

Balls Rearrangement

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

Total Submission(s): 322    Accepted Submission(s): 114


[align=left]Problem Description[/align]
 

　　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.
 

 

[align=left]Input[/align]
 

　　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).
 

 

[align=left]Output[/align]
 

　　For each test case, output the total cost.

 

思路：我们可以发现循环节为a,b的最小公倍数tmp，所以我们只要求从1到tmp的花费即可，易知a和b的大小关系对题目答案没有影响，不妨设a>b，设从1到tmp的花费依次为 w1,w2,w3,....wtmp，我们可以将这tmp个值分成每b个一组，一共a/gcd组（gcd为a和b的最大公约数）,我们设num[i]为第i组的和，现在的关键问题是num[i]怎么求，对于每一组num[i]，这时第一个数在a盒子中的第po个，那么它的花费就是po-1（因为这个数一定是在b盒子的第一个），那么对于后面的第2个，第三个。。。。。。的花费也将是po-1，直到走到第po+b-1个，或者到第a个盒子（其实就是a和po+b-1去个小）,若走到了第po+b-1个盒子，则这一组已经算完，为b*(po-1)花费，且po将被更新为po+b，否则，还应该计算剩下来的花费，这个和前一半类似，只不过这是a盒子从1开始，b盒子从a-po+2开始而已，这也是一串相同的花费，可以O(1)求出，然后更新po。将所有组求完后，剩下的工作就简单了。这里不再罗嗦。

以下是代码。

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define ll long long
using namespace std;
ll dp[100010],sum[100010];
int Po[100010];
long long GCD(long long a,long long b)
{
if(b==0)
return a;
return GCD(b,a%b);
}
ll getans(ll x)
{
if(x>0)
return x;
return -x;
}
int main()
{
//freopen("dd.txt","r",stdin);
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
memset(dp,0,sizeof(dp));
memset(sum,0,sizeof(sum));
ll n,a,b;
cin>>n>>a>>b;
if(a<b)
swap(a,b);
if(a==b)
printf("0\n");
else
{
ll tmp=GCD(a,b);
ll tt=tmp;
tmp=a*b/tmp;
dp[1]=0;
sum[1]=0;
ll po=b+1;
Po[1]=1;
Po[2]=b+1;
for(int i=2;i<=a/tt;i++)
{
if(po+b-1<=a)
{
dp[i]=(po-1)*b;
po=po+b;
if(po>a)
po-=a;
}
else
{
dp[i]=(po-1)*(a-po+1);
int tt=a-po+2;
dp[i]+=(tt-1)*(b-tt+1);
po=b-tt+2;
if(po>a)
po-=a;
}
Po[i+1]=po;
sum[i]=sum[i-1]+dp[i];
}

ll ans=0;
ans=sum[a/tt]*(n/tmp);
n%=tmp;
// cout<<n<<endl;
ll x=n/b;
ans+=sum[x];
n%=b;
if(n)
{
ll PO=Po[x+1];
for(int i=1;i<=n;i++)
{
ans+=getans(PO-i);
PO++;
if(PO>a)
PO-=a;
}
}
cout<<ans<<endl;
}

}
return 0;
}