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 -> llreturn 0;
}