HDU1717 (repeating decimal fraction)

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=1717

Analysis:

We divide this repeating decimal into three parts 0.a1a2a3a4....an (B1B2BB3...BM) into 0, non-cyclic part a1a2...an, and loop section B1B2B3...BM

Set this decimal for X,A1A2A3....AN=S1, b1b2b3...bm=s2,y=0. (B1B2B3B4. Bm

X * 10^n = s1 + y; ---------1)

Y * 10^m = s2 + y; ----------2)

by 2) y = s2/(10^m-1); ------------3)

1), 3) United

x= (s2 + s1 * (10^m-1))/(10^n * (10^m-1));

Pay attention to m=0, or n=0, when dealing with the situation

The code is as follows:

#include <iostream> #include <cmath> #include <cstdio> #include <cstring>using namespace std; typedef long Long LL;    ll GCD (ll A,ll b) {if (b) return gcd (b,a%b); return A;}    Char A[10000];int main () {int n;    scanf ("%d", &n);        while (n--) {scanf ("%s", a);        LL z1 = 0,cnt=0,m=0,cnt2=0,ff=100000,z2=0; for (int i=1;i<strlen (a); i++) {if (a[i]!= '. ')                &&a[i]!= ' (' &&i<ff) {z1=z1*10+a[i]-' 0 ';            if (I&LT;FF) cnt++;            } if (a[i]== ' (') ff=i;        if (i>ff&&a[i]!= ') ') z2=z2*10+a[i]-' 0 ', cnt2++;        } LL tmp1=1,tmp2=1,tmp,z;        for (int i=0;i<cnt;i++) tmp1*=10;        for (int i=0;i<cnt2;i++) tmp2*=10;            if (tmp1==1) {z=z2;            m=tmp2-1;            TMP=GCD (Z,M);            printf ("%i64d/%i64d\n", z/tmp,m/tmp);        Continue     } if (tmp2==1) {z=z1,m=tmp1;       TMP=GCD (Z,M);            printf ("%i64d/%i64d\n", z/tmp,m/tmp);        Continue        } z=z2+ (tmp2-1) *z1;        M=tmp1* (tmp2-1);        TMP=GCD (Z,M);    printf ("%i64d/%i64d\n", z/tmp,m/tmp); } return 0;}

HDU1717 (repeating decimal fraction)