hdu-5728 PowMod(数论)

题目链接：

PowMod

Time Limit: 3000/1500 MS (Java/Others)

Memory Limit: 262144/262144 K (Java/Others)


[align=left]Problem Description[/align]
Declare:
k=∑mi=1φ(i∗n) mod 1000000007

n is a square-free number.

φ is the Euler's totient function.

find:
ans=kkkk...k mod p

There are infinite number of k

[align=left]Input[/align]
[align=left] [/align]
Multiple test cases(test cases ≤100), one line per case.

Each line contains three integers, n,m and p.

1≤n,m,p≤10^7

[align=left]Output[/align]
[align=left] [/align]
For each case, output a single line with one integer, ans.

[align=left]Sample Input[/align]
[align=left] [/align]

1 2 6

1 100 9

[align=left]Sample Output[/align]
[align=left] [/align]

4

7

题意：

先算出那个k的值，再根据指数循环节算出答案;

思路：

这个博客给出了具体的推导过程,我就是参考这个的,而且代码也是;
传送门

AC代码;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e7+10;
const int maxn=500+10;
const double eps=1e-8;

int phi
,vis
,prime
,cnt;
LL sum
,a[100];
inline void Init()
{
cnt=0;
sum[1]=1;
phi[1]=1;
For(i,2,N-1)
{
if(!vis[i])
{
for(int j=2*i;j<N;j+=i)
{
if(!vis[j])phi[j]=j;
vis[j]=1;
phi[j]=phi[j]/i*(i-1);
}
phi[i]=i-1;
prime[++cnt]=i;
}
sum[i]=(sum[i-1]+phi[i])%mod;
}
}

LL pow_mod(LL x,LL y,LL mo)
{
LL s=1,base=x;
while(y)
{
if(y&1)s=s*base%mo;
base=base*base%mo;
y>>=1;
}
return s;
}

LL work(LL a,LL b)
{
if(b==1)return 0;
LL sum=work(a,phi[b]);
sum=sum+phi[b];
LL ans=pow_mod(a,sum,b);
return ans;
}

LL dfs(int pos,LL n,LL m)
{
if(n==1)return sum[m];
if(m==0)return 0;
return ((a[pos]-1)*dfs(pos-1,n/a[pos],m)%mod+dfs(pos,n,m/a[pos]))%mod;
}

inline LL solve(LL n,LL m)
{
int num=0;
LL temp=n;
if(!vis
)a[++num]=n;
else
{
for(int i=1;i<=cnt;i++)
{
if(n<prime[i])break;
if(n%prime[i]==0)
{
a[++num]=prime[i];
n/=prime[i];
}
}
}
return dfs(num,temp,m);
}

int main()
{
Init();
LL n,m,p;
while(cin>>n>>m>>p)
{
LL k=solve(n,m);
LL ans=work(k,p);
print(ans);
}

return 0;
}