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。
 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 = ; i * i <= n; i++) {
if(n % i == ) return false;
}
return true;
} bool fun(int a, int b) //判断两整数奇偶是否相同
{
return !((a & ) & (b & ));
} void solve()
{
int i, j, t;
long long count = ;
for(i = ; i <= ; i++) {
if(i % && prim(i)) {
count += i - ;
continue;
}
count++;
for(j = ; j < i; j++) {
if(fun(i,j) && (i % j != )) {
t = gcd(i, j);
if(t == ) count++; //如果i,j互素,符合
}
}
}
printf("%lld\n", count);
} int main()
{
solve();
return ;
}

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 = ; i * i <= n; i++) {
if(n % i == ) return false;
}
return true;
} bool fun(int a, int b) //判断两整数奇偶是否相同
{
return !((a & ) & (b & ));
} void solve()
{
int i, j, t;
long long count = ;
for(i = ; i <= N; i += ) {
count++;
for(j = ; j < i; j += ) {
t = gcd(i, j);
if(t == ) count++;
} }
for(i = ; i < N; i += ) {
count++;
if(prim(i)) {
count += i - ;
continue;
} else {
for(j = ; j < i; j++) {
if(gcd(i, j)) count++;
}
} }
printf("%lld\n", count);
} int main()
{
solve();
return ;
}

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 = ; i < N; i++) {
a[i] = true;
}
M = (int)sqrt(N);
for(i = ; 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 / ;
i = ;
count = n - ;
while(i <= t) {
if(a[i]) {
if(n % i == ) {
count--;
for(j = i; j * i < n; j++) {
count--;
}
}
}
i++;
}
return count;
} int main()
{
int i;
long long count = ; Eratosthenes(); for(i = ; i < N; i++) {
if(a[i]) {
count += i - ;
} else {
count += fun(i);
}
}
printf("%lld\n", count); return ;
}

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: acutus@126.com)
// 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 = ;
memset(phi, , sizeof(phi)) ;
phi[] = ;
for(i = ; 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 - );
}
}
}
} void solve()
{
int i;
long long ans =;
for(i = ; i < N; i++) {
ans += phi[i];
}
printf("%lld\n", ans);
} int main()
{
genPhi(N);
solve();
return ;
}
Answer:
303963552391

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