poj 2154 color

Color

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 10383 Accepted: 3378

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

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【分析】
低头！http://blog.csdn.net/acdreamers/article/details/8656247
快速幂写错了！快速幂写错了！快速幂写错了！
白学了！退役了！

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【代码】

//poj 2154 color #include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define ll long long#define M(a) memset(a,0,sizeof a)#define fo(i,j,k) for(int i=j;i<=k;i++)using namespace std;const int mxn=36005;bool vis[mxn];int pri[mxn];int T,n,p,tot;inline void init(){    fo(i,2,36000)    {        if(!vis[i]) pri[++tot]=i;        for(int j=1;j<=tot && i*pri[j]<=36000;j++)        {            vis[i*pri[j]]=1;            if(i%pri[j]==0) break;        }    }}inline int phi(int x){    int cnt=x;    for(int i=1;pri[i]*pri[i]<=x;i++)    {        if(x%pri[i]==0)        {            cnt-=cnt/pri[i];            while(x%pri[i]==0) x/=pri[i];        }    }    if(x>1) cnt-=cnt/x;    return cnt%p;}inline int ksm(int x,int k){    x=x%p;    if(k==0) return 1;    if(k==1) return x;    int tmp=ksm(x,k>>1);    if(k&1) return tmp*tmp%p*x%p;    return tmp*tmp%p;}int main(){    init();    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+phi(n/i)*ksm(n,i-1))%p;              if(i*i==n) continue;              ans=(ans+phi(i)*ksm(n,n/i-1))%p;          }        printf("%d\n",ans);    }    return 0;}