Color
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 7630 | Accepted: 2507 |
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.
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
解题思路：
转载：http://blog.csdn.net/tsaid/article/details/7366708
题意：给出两个整数n和p，代表n个珠子，n种颜色，要求不同的项链数，翻转置换不考虑。
结果模p.
题解：
我们知道gcd（i,n)表示了循环节的个数。
比如gcd(2,6) = 2, 它的详细过程为：[1。3。5] [2。4，6]
对于随意一个循环置换，他全部循环节的长度为 n / gcd(i,n)，在上面的样例中: 循环节长度 = 6 / gcd(2,6) = 3
为了方便说明。用L表示循环节的长度，显然 L | n
假设我们枚举L，求出对于每个L有多少个i, 使得 L = n / gcd (i,n)， 那么我们实际上也得到了循环节个数为 n / L 的置换个数。
将L = n / gcd (i,n)转换一下得到：n / L = gcd（i,n )
设 cnt ＝ n / L = gcd(i, n) 注：cnt表示循环节个数，L表示每个循环节的长度
由于 cnt | i, 所以可令 i ＝ cnt * t; ( 由于0 <= i < n, 所以0 <= t < n / cnt = L )
又由于 cnt = n / L， 所以 n = cnt * L;
则 gcd ( i, n ) = gcd ( cnt*t, cnt*L ) = cnt; ①
可知当 gcd ( t, L ) = 1 时 ① 式成立。
因为 gcd ( t, L ) = 1 的个数就是 Euler(L)的个数。
所以我们能够得到结论：循环节个数为n/L的置换有Euler(L)个。
代码：
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
bool isprime[50001];
int prime[50001];
int len=0;;
int n,p; void sieve()
{
for(int i=0;i<=50000;i++)
isprime[i]=1;
isprime[0]=isprime[1]=0;
for(int i=2;i<=50000;i++)
{
if(isprime[i])
{
prime[len++]=i;
for(int j=2*i;j<=50000;j+=i)
isprime[j]=0;
}
}
} int euler(int n)
{
int res=n;
for(int i=0;i<len&&prime[i]*prime[i]<=n;i++)
{
if(n%prime[i]==0)
{
res=res/prime[i]*(prime[i]-1);
while(n%prime[i]==0)
n/=prime[i];
}
}
if(n>1)
res=res/n*(n-1);
return res;
} int pow(int p,int n,int mod)
{
int ans=1;
p=p%mod;
while(n)
{
if(n&1)
ans=ans*p%mod;
p=p*p%mod;
n/=2;
}
return ans;
} int main()
{
sieve();
int t;
scanf("%d",&t);
while(t--)
{
int ans=0;
scanf("%d%d",&n,&p);
for(int i=1;i*i<=n;i++)
if(n%i==0)
{
ans=(ans+euler(i)%p*pow(n,n/i-1,p))%p;//注意取余的位置。 euler(i)%p不取余就WA
if(i*i==n)//仅仅要一个i就能够了
break;
ans=(ans+euler(n/i)%p*pow(n,i-1,p))%p;
}
printf("%d\n",ans);
}
return 0;
}