（Problem 72）Counting fractions
Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d 8 in ascending order of size, we get:
1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
It can be seen that there are 21 elements in this set.
How many elements would be contained in the set of reduced proper fractions for d 1,000,000?
题目大意：
考虑分数 n/d, 其中n 和 d 是正整数。如果 nd 并且最大公约数 HCF(n,d)=1, 它被称作一个最简真分数。
如果我们将d 8的最简真分数按照大小的升序列出来，我们得到：
1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5 , 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
可知该集合中共有21个元素。
d 1,000,000的最简真分数集合中包含多少个元素？
算法一（超时）：
优化技巧：
1、当分母n是素数，则以n为分母的最简真分数的数目为n-1
2、当分母和分子奇偶不同的时候，进一步检查分母和分子的最大公约数
#include<stdio.h> #include<stdbool.h> void swap(int* a, int* b) //交换两值的函数 { int t; t = *a; *a = *b; *b = t; } int gcd(int a, int b) //求最大公约数函数 { int r; if(a < b) swap(&a, &b); while(b) { r = a % b; a = b; b = r; } return a; } bool prim(int n) //判断素数函数 { int i; for(i = 2; i * i <= n; i++) { if(n % i == 0) return false; } return true; } bool fun(int a, int b) //判断两整数奇偶是否相同 { return !((a & 1) & (b & 1)); } void solve() { int i, j, t; long long count = 0; for(i = 2; i <= 100000; i++) { if(i % 2 && prim(i)) { count += i - 1; continue; } count++; for(j = 2; j < i; j++) { if(fun(i,j) && (i % j != 0)) { t = gcd(i, j); if(t == 1) count++; //如果i,j互素，符合 } } } printf("%lld\n", count); } int main() { solve(); return 0; }
算法二：
将1~1000000的整数分奇偶两部分计算，依然超时
#include<stdio.h> #include<stdbool.h> #define N 1000000 int gcd(int a, int b) //求最大公约数函数 { int r; while(b) { r = a % b; a = b; b = r; } return a; } bool prim(int n) //判断素数函数 { int i; for(i = 2; i * i <= n; i++) { if(n % i == 0) return false; } return true; } bool fun(int a, int b) //判断两整数奇偶是否相同 { return !((a & 1) & (b & 1)); } void solve() { int i, j, t; long long count = 0; for(i = 2; i <= N; i += 2) { count++; for(j = 3; j < i; j += 2) { t = gcd(i, j); if(t == 1) count++; } } for(i = 3; i < N; i += 2) { count++; if(prim(i)) { count += i - 1; continue; } else { for(j = 2; j < i; j++) { if(gcd(i, j)) count++; } } } printf("%lld\n", count); } int main() { solve(); return 0; }
算法三：
#include<stdio.h> #include<stdbool.h> #include<math.h> #define N 1000001 bool a[N]; void Eratosthenes() { int i, M, j; for(i = 2; i < N; i++) { a[i] = true; } M = (int)sqrt(N); for(i = 2; i <= M; i++) { if(a[i]) { j = i * i; for(; j < N; j += i) a[j] = false; } } } int fun(int n) { int t, i, count, j; t = n / 2; i = 2; count = n - 1; while(i <= t) { if(a[i]) { if(n % i == 0) { count--; for(j = i; j * i < n; j++) { count--; } } } i++; } return count; } int main() { int i; long long count = 0; Eratosthenes(); for(i = 2; i < N; i++) { if(a[i]) { count += i - 1; } else { count += fun(i); } } printf("%lld\n", count); return 0; }
算法四：使用欧拉函数
//（Problem 72）Counting fractions // Completed on Tue, 18 Feb 2014, 14:08 // Language: C11 // // 版权所有（C）acutus (mail: [email protected]) // 博客地址：http://www.cnblogs.com/acutus/ #include<stdio.h> #include<math.h> #include<stdlib.h> #include<stdbool.h> #define N 1000001 int phi[N]; //数组中储存每个数的欧拉数 void genPhi(int n)//求出比n小的每一个数的欧拉数(n-1的) { int i, j, pNum = 0 ; memset(phi, 0, sizeof(phi)) ; phi[1] = 1 ; for(i = 2; i < n; i++) { if(!phi[i]) { for(j = i; j < n; j += i) { if(!phi[j]) phi[j] = j; phi[j] = phi[j] / i * (i - 1); } } } } void solve() { int i; long long ans =0; for(i = 2; i < N; i++) { ans += phi[i]; } printf("%lld\n", ans); } int main() { genPhi(N); solve(); return 0; }
|
Answer:
|
303963552391 |