Color
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 7693 | Accepted: 2522 |
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
题目大意：
T组测试数据，每组一个n表示1个项链有n个颜色可以涂在n个钻石上，通过旋转相同的算一种方案，问你方案数是多少？
解题思路：
很裸的波利亚计数，转化为的公式就是 ans=sum{ n^( gcd(1,n)-1 ) ,n^( gcd(2,n)-1 ),n^( gcd(3,n)-1 ) .....n^( gcd(n,n)-1 ) }，因为这个n比较大10^9，所以暴力超时。
因此枚举 gcd(k,n)=l 的有多少个，也就是 k=l*x ，n=l*y，也就是gcd(x,y)=1,也就是找到 1~y与y互质的数有多少个，答案：欧拉函数
解题代码：
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
int n,p;
inline ll getPhi(int x){
ll ans=x;
for(int i=2;i*i<=x;i++){
if(x%i==0){
ans=ans/i*(i-1);
while(x%i==0) x/=i;
}
}
if(x>1) ans=ans/x*(x-1);
return ans;
}
inline ll pow_mod(ll a,ll b){
ll sum=1;
while(b>0){
if(b&1) sum=(sum*a)%p;
a=(a*a)%p;
b/=2;
}
return sum%p;
}
int main(){
int t;
scanf("%d",&t);
while(t-- >0){
scanf("%d%d",&n,&p);
ll ans=0;
for(int i=1;i*i<=n;i++){
if(n%i==0){
if(i*i==n) ans=(ans+(getPhi(i)*pow_mod(n,i-1))%p)%p;
else{
ans=(ans+(getPhi(n/i)*pow_mod(n,i-1))%p)%p;
ans=(ans+(getPhi(i)*pow_mod(n,n/i-1))%p)%p;
}
}
}
printf("%lld\n",ans%p);
}
return 0;
}
POJ 2154 Color（组合数学-波利亚计数,数论-欧拉函数,数论-整数快速幂）