Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

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Problem B: Dead Fraction

Mike is frantically scrambling to finish his thesis at the last minute. He needs to assemble all his research notes into vaguely coherent form in the next 3 days. Unfortunately, he notices that he had been extremely sloppy in his calculations. Whenever he needed to perform arithmetic, he just plugged it into a calculator and scribbled down as much of the answer as he felt was relevant. Whenever a repeating fraction was displayed, Mike simply reccorded the first few digits followed by "...". For instance, instead of "1/3" he might have written down "0.3333...". Unfortunately, his results require exact fractions! He doesn't have time to redo every calculation, so he needs you to write a program (and FAST!) to automatically deduce the original fractions.

To make this tenable, he assumes that the original fraction is always the simplest one that produces the given sequence of digits; by simplest, he means the the one with smallest denominator. Also, he assumes that he did not neglect to write down important digits; no digit from the repeating portion of the decimal expansion was left unrecorded (even if this repeating portion was all zeroes).

There are several test cases. For each test case there is one line of input of the form "0.dddd..." where dddd is a string of 1 to 9 digits, not all zero. A line containing 0 follows the last case. For each case, output the original fraction.

Note that an exact decimal fraction has two repeating expansions (e.g. 1/5 = 0.2000... = 0.19999...).

Sample Input

0.2...
0.20...
0.474612399...
0

Output for Sample Input

2/9
1/5
1186531/2500000

---

此題給一個循環小數，但沒給你後面的循環長度，自行劃分找到一個分數表示法產生這個表示法，
輸出一個分母最小的分數。


#include <stdio.h>
#define LL long long
LL gcd(LL x, LL y) {
    LL t;
    while(x%y)
        t = x, x = y, y = t%y;
    return y;
}
LL mpow(LL x, LL y) {
    if(y == 0)  return 1;
    if(y&1) return x*mpow(x*x, y>>1);
    return mpow(x*x, y>>1);
}
int main() {
    char s[105];
    int i;
    while(scanf("%s", s) == 1) {
        if(s[1] == '\0')
            break;
        for(i =  2; s[i]; i++)
            if(s[i] == '.')
                break;
        s[i] = '\0';
        int a = i-2, t;
        LL x, y, ax, ay, g, tx, ty;
        y = mpow(10, a)-1;
        sscanf(s+2, "%lld", &x);
        g = gcd(x, y);
        ax = x/g, ay = y/g;
        for(i = 2; i < a+1; i++) {
            t = s[i+1];
            s[i+1] = '\0';
            sscanf(s+2, "%lld", &x);
            s[i+1]  = t;
            sscanf(s+i+1, "%lld", &y);
            ty = mpow(10, i-1)*(mpow(10, a-i+1)-1);
            tx = x*(mpow(10, a-i+1)-1)+y;
            g = gcd(tx, ty);
            tx /= g, ty /= g;
            if(ty < ay) ax = tx, ay = ty;
            //printf("%lld %lld %d %d %lld/%lld\n", x, y, i-1, a-i+1, tx, ty);
        }
        printf("%lld/%lld\n", ax, ay);
    }
    return 0;
}