solution1

直观上的分数处理

#include <iostream>
using namespace std;
int main()
{
  printf("1048575/524288");
  return 0;
}

#include<stdio.h>
#include<math.h>
typedef long long ll;
struct fraction{
	ll up, down;
};
ll gcd(ll a, ll b){
	if(!b) return a;
	return gcd(b, a % b);
}
fraction r(fraction f){
	if(gcd(f.down, f.up) > 1){
		f.down /= gcd(f.down, f.up);
		f.up /= gcd(f.down, f.up);
	}
	return f;
}
fraction add(fraction f1, fraction f2){
	fraction f;
	f.down = f1.down * f2.down;
	f.up = f1.up * f2.down + f2.up * f1.down;
	return r(f);
}
int main(){
	fraction f, t;
	f.up = f.down = 1;
	for(ll i = 2; i <= pow(2, 19); i *= 2){
		t.up = 1;
		t.down = i;
		printf("%lld %lld\n", t.down, f.up);
		f = add(f, t);
	}
	printf("%lld %lld, %lld %lld", f.up / f.down, f.up % f.down, f.up, f.down);
	return 0;
}

solution2

手动通分计算为
(2¹⁹+2¹⁸+2¹⁷……+2⁰)/2¹⁹= (2²⁰-1)/2¹⁹

• 2⁰+2¹+2²+……+2^n-1 = 2ⁿ-1
  
• 较大的数若比 较小的数 的两倍大于或者小1，则两者互质

#include<stdio.h>
#include<math.h>
typedef long long ll;
int main(){
	printf("%lld/%lld", (ll) pow(2, 20) - 1, (ll) pow(2, 19));//注意别漏了强转double -> ll
	return 0;
}