标签：
Goffi and GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 992 Accepted Submission(s): 336
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
Sample Output
2
1
Hint
For the first case, (2, 1) and (1, 2) satisfy the equality.
Source
题意:求使得
成立的(a,b)的个数.
首先,我们可以知道 gcd(n,a)<=n,所以当 k>2 的时候没有这样的(a,b)
然后当 k==2 的时候我们只有一个这样的组合 (a,b) = (n,n)
接下来考虑 k=1的情况:当 k = 1时 ,gcd(n-a,n) = gcd(a,n) (这里源自gcd的变换公式) 整个式子就转换成了 gcd(a,n)*gcd(b,n)=n
设 gcd(a,n) 为 x,那么 gcd(a/x,n/x)=1，a/x 与 n/x 互质，利用欧拉函数 phi(x) 可以得到 a的个数，同理可以得到b的个数.并且x也是 n 的因子，所以在求解n的因子的同时就可以将(a,b)求出来了。
这里总结一个公式：如果d是n的一个约数，那么1<=i<=n中 gcd(i,n) = d的个数是phi(n/d)，即n/d的欧拉函数
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
typedef long long LL;
const LL mod = 1000000007;
LL phi(LL x)
{
LL ans=x;
for(LL 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;
}
LL n,k;
int main()
{
while(scanf("%lld%lld",&n,&k)!=EOF)
{
if(n==1) {
printf("1\n");
continue;
}
if(k>1)
{
if(k==2)
printf("1\n");
else printf("0\n");
continue;
}
LL sum = 0;
for(LL i=1; i*i<=n; i++)
{
if(n%i==0)
{
if(i*i==n) sum = (sum+phi(i)*phi(i))%mod;
else {
sum = (sum+2*phi(i)*phi(n/i))%mod;
}
}
}
printf("%lld\n",sum);
}
return 0;
}
hdu 4983(欧拉函数)
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原文地址：http://www.cnblogs.com/liyinggang/p/5630987.html
