hdoj4893Goffi and GCD【欧拉函数】

Goffi and GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 951    Accepted Submission(s): 319


Problem Description

Goffi is doing his math homework and he finds an equality on his text book: gcd(n−a,n)×gcd(n−b,n)=nk.

Goffi wants to know the number of (a,b)
satisfy the equality, if n and k are
given and 1≤a,b≤n.

Note: gcd(a,b) means
greatest common divisor of a and b.

 

Input

Input contains multiple test cases (less than 100). For each test case, there's one line containing two integers n and k (1≤n,k≤109).

 

Output

For each test case, output a single integer indicating the number of (a,b)
modulo 109+7.

 

Sample Input

2 1
3 2

 

Sample Output

2
1
Hint
For the first case, (2, 1) and (1, 2) satisfy the equality.

 

Source

BestCoder Round #6

 

题意： 给你 N 和 K,问有多少个数对满足 gcd(N-A, N) * gcd(N - B, N) = N^K

分析： 由于 gcd(a, N) <= N,于是 K>2 都是无解,K=2 只有一个解 A=B=N,只要考虑 K = 1 的情况就好了 其实上式和这个是等价的 gcd(A, N) * gcd(B, N) = N^K,我们枚举 gcd(A, N) = g,那么gcd(B, N) = N / g。问题转化为统计满足 gcd(A, N) = g 的 A 的个数。这个答案就是 ɸ(N/g) 只要枚举 N 的 约数就可以了。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#define MOD 1000000007
using namespace std;
const int maxn=10010;
long long Euler(long long  n){
long long  ret=1,i;
for(i=2;i*i<=n;++i){
if(n%i==0){
n/=i;ret*=i-1;
while(n%i==0){
n/=i;ret*=i;
}
}
}
if(n>1)ret*=n-1;
return ret;
}
int main()
{
long long n,k,i,j;
while(scanf("%lld%lld",&n,&k)!=EOF){
if(n==1||k==2){
printf("1\n");
}
else if(k>2){
printf("0\n");
}
else {
long long ans=0;
for(i=1;i*i<=n;++i){
if(n%i==0){
long long num=Euler(i)*Euler(n/i)%MOD;
if(i*i==n){
ans+=num;
}
else {
ans=ans+num*2;
}
ans%=MOD;
}
}
printf("%lld\n",ans);
}
}
return 0;
}