2 条题解
- 
  3njnu+19180232 LV 7 @ 2018-12-25 15:34:04 #include <stdio.h> int f(int a,int b); 
 int main()
 {
 int x;scanf("%d",&x);
 int count=0;
 for(int i=2;i<x;i++)
 {
 int ii=i;
 for(int j=ii+1;j<=x;j++)
 if(f(ii,j)==-1) count++;
 }
 printf("%d",count);
 }int f(int a,int b) 
 {
 int t;
 if(a < b)
 {
 t = a;a = b;b = t;
 }
 while(t=a%b)
 {
 a = b;b = t;
 }
 if(b == 1)
 return -1;
 return 0;
 }
- 
  1@ 2023-12-26 15:04:59#include<iostream> 
 using namespace std;int gcd(int a,int b) 
 {
 if (b == 0)return a;
 else return gcd(b, a % b);
 }int main() 
 {
 //(2,3), (2,5), (3,4), (3,5), (4,5), (5,6)观察知j>i
 int n; cin >> n;
 int cnt = 0;
 for (int i = 2; i <= n; i++)
 for (int j = i+1; j <= n; j++)
 if (gcd(i, j) == 1 )cnt++;//3 4 4 3 3 5 5 3 重复计算cout << cnt; 
 return 0;
 }
- 1
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