Ultraviolet A 10555-dead fraction (number theory + infinite repeating decimals)

Ultraviolet A 10555-dead Fraction

Question Link

Given a repeating decimal point, the unit is uncertain. The decimal point is represented by a fraction with the minimum denominator.

Idea: Push a small Formula
A decimal point 0. aaaaabbb... is in the form of N/m, and a is an integer with a C digit, and B is a decimal part with a D digit.
So AAAAA. BBB... And aaaaabbb can be expressed as 10 respectively.C? (N/M) And 10C+D? (N/M)
The two methods are subtracted:AAAAABBB?AAAAA= (10C+D? 10C)(N/M)
Set n1 = aaaaabbb, n2 = AAAAA, M1 = 10C+D, M2 = 10C.
Therefore, N/m can be expressed as (N1-N2)/(M1-m2)

To solve this problem, calculate the positions of N1, N2, M1, and m2 respectively to indicate the scores and record the minimum denominator values.

Code:

#include <stdio.h>#include <string.h>char str[105];const long long INF = 0x3f3f3f3f3f3f3f;long long gcd(long long a, long long b) {    if (!b) return a;    return gcd(b, a % b);}void solve() {    int len = strlen(str) - 5;    for (int i = 0; i < len; i++)str[i] = str[i + 2];    long long n1 = 0, m1 = 1;    long long anszi, ansmu = INF;    for (int i = 0; i < len; i++) {n1 = n1 * 10 + str[i] - '0';m1 *= 10;    }    for (int i = 0; i < len; i++) {int n2 = 0, m2 = 1;for (int j = 0; j < i; j++) {    n2 = n2 * 10 + str[j] - '0';    m2 *= 10;}long long zi = n1 - n2, mu = m1 - m2;long long k = gcd(zi, mu);zi /= k; mu /= k;if (mu < ansmu) {    anszi = zi;    ansmu = mu;}    }    printf("%lld/%lld\n", anszi, ansmu);}int main() {    while (~scanf("%s", str) && strcmp(str, "0") != 0) {solve();    }    return 0;}