POJ2154——Color(Polya定理+筛素数+欧拉函数)
2013-05-10 21:13
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Color
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 6292
Accepted: 2090
Description
Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all the N colors, and the repetitions
that are produced by rotation around the center of the circular necklace are all neglected.
You only need to output the answer module a given number P.
Input
The first line of the input is an integer X (X <= 3500) representing the number of test cases. The following X lines each contains two numbers N and P (1 <= N <= 1000000000, 1 <= P <= 30000), representing a test case.
Output
For each test case, output one line containing the answer.
Sample Input
5
1 30000
2 30000
3 30000
4 30000
5 30000
Sample Output
1
3
11
70
629
Source
POJ Monthly,Lou Tiancheng
解析:
Poj2154的N比较大,10^9的范围,但是只有旋转这种置换,没有翻转。
如果我们按照O(n)的做法,即∑m^gcd(n,i),显然是要超时的,所以需要换一种思路来计算。
设循环节长度为a=gcd(n,i),则一个循环的长度为L=n/a。由以上两式可以得到gcd(L*a,k*a)=a,其中k为不超过L的整数。进一步化简得到gcd(L,k)=1,那么满足这个式子的 循环的个数,也就是k的个数,就是euler(L),euler代表欧拉函数(小于L且与L互质的数的个数)。
所以答案就是(∑euler(L)*m^(n/L)) / n,本题中m=n,上式也就是(∑euler(L)*n^(n/L-1)。因此只需要枚举n的约数L,L从1枚举到sqrt(n)即可,因为另一个约数就是n/L。注意 当L*L=n的时候别重复算。时间复杂度O(n^0.5logn)。
代码:
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int n,mod,test,m=0;
bool col[50000];
int a[32000];
void prime(int n)
{
for(int i=2;i<=n;i++)
{
if(!col[i])
{
a[++m]=i;
for(int j=i*i;j<=n;j+=i)col[j]=true;
}
}
}
int quick(int a,int b)
{
int tmp=1;
for(a%=mod;b;b>>=1)
{
if(b&1)tmp=tmp*a%mod;
a=a*a%mod;
}
return tmp;
}
int euler(int n)
{
int tmp=n;
for(int i=1;i<=m,a[i]*a[i]<=n;i++)
if(n%a[i]==0)
{
tmp=tmp/a[i]*(a[i]-1);
while(n%a[i]==0)n/=a[i];
}
if(n>1)tmp=tmp/n*(n-1);
return tmp%mod;
}
int main()
{
freopen("poj2154.in","r",stdin);
freopen("poj2154.out","w",stdout);
scanf("%d",&test);
prime(32000);
while(test--)
{
int ans=0,i;
scanf("%d%d",&n,&mod);
for(i=1;i*i<n;i++)
if(n%i==0)ans=(ans+euler(i)*quick(n,n/i-1)+euler(n/i)*quick(n,i-1))%mod;
if(i*i==n)ans=(ans+euler(i)*quick(n,i-1))%mod;
printf("%d\n",ans);
}
return 0;
}
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 6292
Accepted: 2090
Description
Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all the N colors, and the repetitions
that are produced by rotation around the center of the circular necklace are all neglected.
You only need to output the answer module a given number P.
Input
The first line of the input is an integer X (X <= 3500) representing the number of test cases. The following X lines each contains two numbers N and P (1 <= N <= 1000000000, 1 <= P <= 30000), representing a test case.
Output
For each test case, output one line containing the answer.
Sample Input
5
1 30000
2 30000
3 30000
4 30000
5 30000
Sample Output
1
3
11
70
629
Source
POJ Monthly,Lou Tiancheng
解析:
Poj2154的N比较大,10^9的范围,但是只有旋转这种置换,没有翻转。
如果我们按照O(n)的做法,即∑m^gcd(n,i),显然是要超时的,所以需要换一种思路来计算。
设循环节长度为a=gcd(n,i),则一个循环的长度为L=n/a。由以上两式可以得到gcd(L*a,k*a)=a,其中k为不超过L的整数。进一步化简得到gcd(L,k)=1,那么满足这个式子的 循环的个数,也就是k的个数,就是euler(L),euler代表欧拉函数(小于L且与L互质的数的个数)。
所以答案就是(∑euler(L)*m^(n/L)) / n,本题中m=n,上式也就是(∑euler(L)*n^(n/L-1)。因此只需要枚举n的约数L,L从1枚举到sqrt(n)即可,因为另一个约数就是n/L。注意 当L*L=n的时候别重复算。时间复杂度O(n^0.5logn)。
代码:
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int n,mod,test,m=0;
bool col[50000];
int a[32000];
void prime(int n)
{
for(int i=2;i<=n;i++)
{
if(!col[i])
{
a[++m]=i;
for(int j=i*i;j<=n;j+=i)col[j]=true;
}
}
}
int quick(int a,int b)
{
int tmp=1;
for(a%=mod;b;b>>=1)
{
if(b&1)tmp=tmp*a%mod;
a=a*a%mod;
}
return tmp;
}
int euler(int n)
{
int tmp=n;
for(int i=1;i<=m,a[i]*a[i]<=n;i++)
if(n%a[i]==0)
{
tmp=tmp/a[i]*(a[i]-1);
while(n%a[i]==0)n/=a[i];
}
if(n>1)tmp=tmp/n*(n-1);
return tmp%mod;
}
int main()
{
freopen("poj2154.in","r",stdin);
freopen("poj2154.out","w",stdout);
scanf("%d",&test);
prime(32000);
while(test--)
{
int ans=0,i;
scanf("%d%d",&n,&mod);
for(i=1;i*i<n;i++)
if(n%i==0)ans=(ans+euler(i)*quick(n,n/i-1)+euler(n/i)*quick(n,i-1))%mod;
if(i*i==n)ans=(ans+euler(i)*quick(n,i-1))%mod;
printf("%d\n",ans);
}
return 0;
}
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