[UVA][分循] 10555 - Dead Fraction＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2012-12-23 18:52:40| 人1,691| 回0 | 上一篇 | 下一篇

[UVA][分循] 10555 - Dead Fraction

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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

此一循小，但你後面的循度，自行分找到一分表示法生表示法，
出一分母最小的分。

#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);
nbsp;          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;
}