题目大意:给出一个小数,从...開始能够是不论什么数字,可是保证是无限循环小数。将该小数用分式的形式表示,而且要求分母尽量大。
解题思路:这题主要是怎么将无限循环小数转换成分式,这种:
- 有小数0.abcdEEE,未循环部分长4。循环节为E,E的长度为i(如果)
- abcd+E999…(i位9)10i
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 105;
const ll INF = 0x3f3f3f3f3f3f3f;
char s[maxn];
ll gcd (ll a, ll b) {
return b ?
gcd(b, a%b) : a;
}
int main () {
while (scanf("%s", s) == 1 && strcmp(s, "0")) {
int len = strlen(s)-5;
ll ansu, ansd = INF;
for (int i = 0; i < len; i++)
s[i] = s[i+2];
for (int i = 0; i < len; i++) {
ll d = 1, u = 0;
for (int j = 0; j < i; j++) {
d = d * 10;
u = u * 10 + s[j] - '0';
}
ll x = 0, y = 0;
for (int j = i; j < len; j++) {
x = x * 10 + s[j] - '0';
y = y * 10 + 9;
}
d = d * y;
u = u * y + x;
ll g = gcd(d, u);
u /= g;
d /= g;
if (d < ansd) {
ansd = d;
ansu = u;
}
}
printf("%lld/%lld\n", ansu, ansd);
}
return 0;
}