HDU 2588 GCD && GCD问题总结

                                                                                                             GCD(一)

题目：

 The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common
to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.

   求满足题目要求的x个数。

算法：

   直接筛选会超时，根据题目给出的不等式特点GCD(x,N) >= M 可以知道满足题目要求的一定是N的因子而且必须大于等于M（想想为什么？解体关键）。所以，只要枚举N的大于等于M的因子就可以了。因为，在10^9内最多的因子数不超过30个。所以，总时间是O(30*loglogn)接近常数。

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

typedef __int64 LL;
const int MOD = 1000000007;

int euler_phi(int n){
int k = (int)sqrt(n + 0.5);
int ans = n;
for(int i = 2;i <= k;++i) if(0 == n % i){
ans = ans / i * (i - 1);
while(0 == n % i) n /= i;
}

if(n > 1) ans = ans / n * (n - 1);
return ans;
}

LL getFact(int n,int m){
LL res = 0;
int k = sqrt(n + 0.5);
int tmp;

for(int i = 1;i <= k;++i){
if(0 == n % i){
tmp = n / i;
if(i >= m) res += euler_phi(n / i);
if(tmp >= m && i != tmp) res += euler_phi(n / tmp);
}
}
return res;
}

int main()
{
int T,n,m;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);

if(n == 1 && m == 1){
puts("1");
continue;
}

printf("%I64d\n",getFact(n,m));
}
return 0;
}

 

  

                                GCD(二)

题目：

   给你一个数N,使得在1~N之间能够找到x使得x满足gcd（ x ,  N  ） >= M，求解gcd(x,N)的和。

算法：

  由上题的知识可以知道，1...N的互质个数为欧拉函数值且其gcd只能是N的因子。所以，对于N = x * y。我们只要

求出x在y内的互质个数就好了，结果乘以x就是gcd = x的和了.

证明：

   SUM(gcd = x ) = 1*x + 2*x + 3*x ..... y*x

  所以，当gcd = x的时候只要求出y的欧拉函数值就好了。

 

而一个数的因子又可以在sqrt(N)内求出。

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

typedef long long LL;

int euler_phi(int n){
int m = sqrt(n + 0.5);
int ans = n;
for(int i = 2;i <= m;++i) if(0 == n % i){
ans = ans / i * (i - 1);
while(0 == n % i) n /= i;
}
if(n > 1) ans = ans / n * (n - 1);

return ans;
}

LL solve(int n,int m){
LL res = 0;
int k = sqrt(n + 0.5);
for(int i = 1;i <= k;++i){
if(0 == n % i){
if(i >= m)
res += i * euler_phi(n / i);
if(i != n / i && n / i >= m)
res += n / i * euler_phi(i);
}
}
return res;
}

int main()
{
int n,m;
while(~scanf("%d%d",&n,&m)){
printf("%lld\n",solve(n,m));
}
return 0;
}

 

                                                                                                  GCD(三)

 

题目：

    The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and
b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M,please answer sum of  X satisfies 1<=X<=N and (X,N)>=M.

 

算法：

   跟GCD(一)不同的是这题求得是满足gcd(x,n) >= m ，x的和。而由欧拉函数中的一个定理可以知道


 

所以，只要SUM(n = x * y) = y*α(y) / 2 * x 

因为要的是x的和，而我们是在把X先进行X / x处理的所以最后要在乘回上x得到原值。

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

typedef long long LL;
const int MOD = 1000000007;

int euler_phi(int n){
int k = (int)sqrt(n + 0.5);
int ans = n;
for(int i = 2;i <= k;++i) if(0 == n % i){
ans = ans / i * (i - 1);
while(0 == n % i) n /= i;
}

if(n > 1) ans = ans / n * (n - 1);
return ans;
}

LL getFact(int n,int m){
LL res = 0;
int k = sqrt(n + 0.5);
LL tmp;

for(int i = 1;i <= k;++i){
if(0 == n % i){
tmp = n / i;
if(i >= m){
LL t1 = tmp * euler_phi(tmp) / 2 % MOD;
t1 = t1 ? t1 : 1;
res = (res + t1 * i) % MOD;
}
if(tmp >= m && i != tmp) {
LL t1 = i * euler_phi(i) / 2 % MOD;
t1 = t1 ? t1 : 1;
res = (res + t1 * tmp) % MOD;
}
}
}

return res >= MOD ? res%MOD : res;
}

int main()
{
int T,n,m;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
printf("%lld\n",getFact(n,m));
}
return 0;
}