GCD相关

板子：

 int gcd(int a, int b) { return b >  ? gcd(b, a % b) : a; }

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POJ1930

题意：给你一个无限循环小数，给到小数点后 9 位，要求这个数的分数形式。

解法：

要想解决这道题，首先应该了解如何将循环小数化为分数：

一，纯循环小数化分数：循环节的数字除以循环节的位数个 9 组成的整数。例如： 
$0.3333……=3/9=1/3$;
$0.285714285714……=285714/999999＝2/7$. 
二，混循环小数：（例如：$0.24333333……$）不循环部分和循环节构成的的数减去不循环部分的差，再除以循环节位数个 9 添上不循环部分的位数个  0。例如： 
$0.24333333…………=(243-24)/900=73/300 $
$0.9545454…………=(954-9)/990=945/990=21/22$

这便是循环小数化成分数的方法，知道这个后，解决这道题也好办了。

代码如下：

 #pragma GCC optimize(3)
 #include <time.h>
 #include <algorithm>
 #include <bitset>
 #include <cctype>
 #include <cmath>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <deque>
 #include <functional>
 #include <iostream>
 #include <map>
 #include <queue>
 #include <set>
 #include <sstream>
 #include <stack>
 #include <string>
 #include <utility>
 #include <vector>
 using namespace std;
 const double EPS = 1e-;
 const int INF = ;
 const long long LLINF = ;
 const double PI = acos(-1.0);

 inline int READ() {
     char ch;
     while ((ch = getchar()) <  ||  < ch)
         ;
     int ans = ch - ;
     while ( <= (ch = getchar()) && ch <= )
         ans = (ans << ) + (ans << ) + ch - ;
     return ans;
 }

 #define REP(i, a, b) for (int i = (a); i <= (b); i++)
 #define PER(i, a, b) for (int i = (a); i >= (b); i--)
 #define FOREACH(i, t) for (typeof(t.begin()) i = t.begin(); i != t.end(); i++)
 #define MP(x, y) make_pair(x, y)
 #define PB(x) push_back(x)
 #define SET(a) memset(a, -1, sizeof(a))
 #define CLR(a) memset(a, 0, sizeof(a))
 #define MEM(a, x) memset(a, x, sizeof(a))
 #define ALL(x) begin(x), end(x)
 #define LL long long
 #define Lson (index * 2)
 #define Rson (index * 2 + 1)
 #define pii pair<int, int>
 #define pll pair<LL, LL>
 #define MOD ((int)1000000007)
 #define MAXN 20 + 5
 ///**********************************START*********************************///
 string ss, sn;

 char s[MAXN], ns[MAXN];
 int nex[MAXN];

 void get_next(char* p, int next[]) {
     int pLen = strlen(p) + ;  //要求循环节长度时这里才+1
     next[] = -;
     int k = -;
     int j = ;
     while (j < pLen - ) {
         if (k == - || p[j] == p[k]) {
             ++j;
             ++k;
             if (p[j] != p[k])
                 next[j] = k;
             else
                 next[j] = next[k];
         } else {
             k = next[k];
         }
     }
 }

 int gcd(int a, int b) { return b >  ? gcd(b, a % b) : a; }

 void cal_cal(int len) {
     int up = , down = ;
     for (int i = len - ; i >= ; i--)
         up += int(s[i] - '') * pow(, len - i - );
     for (int i = ; i < len; i++) down +=  * pow(, i);
     int g = gcd(up, down);
     printf("%d/%d\n", up / g, down / g);
     return;
 }

 void cal_cal2(int st, int len) {
     char tmp[];
     int up = , down = ;
     int a, b;

     for (int i = ; i < st; i++) tmp[i] = s[i];
     tmp[st] = '\0';
     a = atoi(tmp);
     for (int i = st; i < st + len; i++) tmp[i - st] = s[i];
     tmp[len] = '\0';
     b = atoi(tmp);

     up = a * pow(, len) + b - a;
     int pos = ;
     for (int i = ; i < len; i++) tmp[pos++] = '';
     for (int i = ; i < st; i++) tmp[pos++] = '';
     down = atoi(tmp);
     int g = gcd(up, down);
     printf("%d/%d\n", up / g, down / g);
     return;
 }

 int main() {
 #ifndef ONLINE_JUDGE
     freopen("input.txt", "r", stdin);
 #endif  // !ONLINE_JUDGE
     while (cin >> ss) {
         if (ss.size() == ) break;
         int pos = ;
         for (int i = ; ss[i] != '.'; i++) s[pos++] = ss[i];
         s[pos] = '\0';
         int flag = ;  // 1为全循环，2为混合循环
         get_next(s, nex);
         int len = pos - nex[pos];
         if (len != pos) {
             cal_cal(len);
         } else {
             int st;
             for (st = ; st < pos; st++) {
                 int nlen = ;
                 for (int i = st; i < pos; i++) {
                     ns[nlen++] = s[i];
                 }
                 CLR(nex);
                 get_next(ns, nex);
                 if (nlen - nex[nlen] != nlen) {
                     cal_cal2(st, nlen);
                     break;
                 } else
                     continue;
             }
         }
     }
     return ;
 }