GCD nyoj 1007 (欧拉函数+欧几里得)

GCD  nyoj 1007 (欧拉函数+欧几里得)

GCD

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

 

描述

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.

 

输入

The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (1<=N<=10^9, 1<=M<=10^9), representing a test case.

输出

Output the answer mod 1000000007

样例输入

3
1 1
10 2
10000 72

样例输出

1
35
1305000


题意：1<= x <= n ，求gcd(x,n) >= m 说有满足条件的x的和  （最后要模Mod=1000000007)
分析：

如果是求满足条件的x的个数：

因为x要满足1<=x<=n 且gcd(x,n)>=m，所以x为n的因子，即gcd(x,n)=x>=m，设y=n/x，
则y的欧拉函数为小于y且与y互质的数的个数。假设与y互质的数为p1,p2,p3……，那么gcd(x*pi,n)=x>=m.
即要找出所有符合要求的y的欧拉函数值的和即可。


因为1<= x <= n 若gcd(x,n) = x >= m 推出x是n的因子  y = n/x 与y互质且小于y的数pi 总共有eular(y)个
gcd(x*pi,n) = x >= m  所以此时满足条件的x的个数有eular(y)个

又因为 gcd(a,n) = gcd(n-a,n) 【定理记住】 所以此条件下x的和sum = n*eular(y)/2
最后再依此求出n的所有因子 计算sum相加即可

计算x的总和：附上代码

/*
author:谦智
HDU 2588-GCD(欧拉函数)  数论 
*/
#include<iostream>
using namespace std;
const int Mod = 1000000007;
#define LL long long
LL eular(LL n) ;
LL eularSum(LL k) {
  if (k == 1) return 1;//特殊值 
  return k*eular(k)/2;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    LL n, m;
    cin >> n >> m;
    LL sum = 0; 
    if (n == 1 && m == 1) {//特殊情况
      cout << 1 << endl;
      continue;
    }
    for (LL i = 1; i*i <= n; i++) {
      if (n%i == 0) {
        if (i >= m) {
          sum = (sum + n*eular(n/i)/2)%Mod; //==》 sum = (sum + i*(n/i)*eular(n/i)/2)%Mod
//          sum = (sum + i*eularSum(n/i))%Mod;//总共有 eular(n/i)个x使得gcd(x,n) >= m 
        } 
        if (n/i >= m && i*i != n) {
　　　　　　//只需要在这里加一个i== 1的情况上面不要加  不然会重复
          if (i == 1) sum = (sum + n)%Mod;//当eular(i) < 2时只需要加上此时唯一满足条件的x即可  其他的eular(i)一定都是偶数因为gcd(a,n)= gcd(n-a,n)
          else sum = (sum + n*eular(i)/2)%Mod;
//           sum = (sum + n/i*eularSum(i))%Mod;
        }
      }
    }
    cout << sum << endl;
  }
} 
LL eular(LL n) {
  LL ans = n;
  for (LL i = 2; i*i <= n; i++) {
    if (n%i == 0) {
      ans = ans/i*(i-1);
      while (n%i == 0) {
        n /= i;
      }
    }
  }
  if (n != 1) ans = ans/n*(n-1);
  return ans;
}

 

计算x的个数：附上代码

 

//计算 x小于n 且gcd(x,n) >= m 的x的个数 
#include<iostream>
using namespace std;
int eular(int n) ;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    int sum = 0; 
    for (int i = 1; i*i <= n; i++) {
      if (n%i == 0) {
        if (i >= m) {
          sum += eular(n/i);
        } else if (n/i >= m && i*i != n) {
          sum += eular(i);
        }
      }
    }
    cout << sum << endl;
  }
} 
int eular(int n) {
  int ans = n;
  for (int i = 2; i*i <= n; i++) {
    if (n%i == 0) {
      ans = ans/i*(i-1);
    }
    while (n%i == 0) {
      n /= i;
    }
  }
  if (n != 1) ans = ans/n*(n-1);
  return ans;
}