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。

49愛的鼓勵 6訂閱站台

Problem A: Moliu Fractions
Time limit: 10 seconds

[Vocabulary] moliu (adj) - (Cantonese) Something is moliu if it is senseless, stupid, boring etc.

As the title suggests, this task is very dull, so... those who hate boring tasks may consider skipping this one. ☺

The reason why this task is so boring is that it is taken from my younger brother's maths homework. He is currently learning "multiplication of fractions", and his teacher has given him loads of related exercises. And... as you know, all the questions in the exercises are so dull in the following aspects:

1. They are unrealistic and meaningless; what's the meaning of finding the weight of 3/8 bar of soap? Or 5/7 cup of cola?
2. The answers are nearly always ugly. Values like 13/24 kilograms are not uncommon.
3. Most important of all, you don't even need to look into the problem statements in order to get the solutions! Just pick out all the numbers stated, multiply them together, and you get the correct answer!!!

For the above reasons, instead of wasting our time in solving such boring questions by hand, why don't we just let the computer to do the work? Please write a program that solves the questions by the method of "pick out numbers then multiply"!

Input and Output

The first line of input contains an integer T. Each of the next T lines is a question to which your program should compute the answer as described above.

This task deals with three types of operands, namely: integers, proper fractions and mixed fractions. As an example, the first question in the sample input reads:

In each question there are at least 2 and at most 4 operands, all of which are well-formatted as shown in sample input. There is always a space (ASCII #32) immediately after each operand. All operands are non-negative, so your result will also be greater than or equal to 0. There are no improper fractions in the input, and your output should also not contain any. Of course, you will have to give your results in the simplest form. If your result can be expressed as an integer, you should not output it as a proper fraction or a mixed fraction. You do not have to worry about integer overflow (after all, the maths exercises are targeted to kids :-).

Sample Input

2
What is the product of 2 , 6/55 and 3-1/3 ?
A cookie weighs exactly 18-2/5 g. What is the weight of a packet of 12 cookies?

Sample Output

8/11
220-4/5

---

Problemsetter: Mak Yan Kei

題目要求把字串內的所有分數抓出來相乘。不用擔心 overflow。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <sstream>
using namespace std;
int gcd(int x, int y) {
    int t;
    while(x%y)
        t = x, x = y, y = t%y;
    return y;
}
int main() {
    int testcase;
    char ss[105];
    int i;
    scanf("%d", &testcase);
    while(getchar() != '\n');
    while(testcase--) {
        string s;
        getline(cin, s);
        for(i = 0; i < s.length(); i++) {
            if(s[i] >= 'a' && s[i] <= 'z')
                s[i] = ' ';
            else if(s[i] >= 'A' && s[i] <= 'Z')
                s[i] = ' ';
        }
        stringstream sin(s);
        int p = 1, q = 1;
        while(sin >> s) {
            if(s[0] < '0' || s[0] > '9')
                continue;
            strcpy(ss, s.c_str());
            int a, b, c;
            if(sscanf(ss, "%d-%d/%d", &a, &b, &c) == 3) {
                b = b+a*c;
                p *= b, q *= c;
            } else if(sscanf(ss, "%d/%d", &b, &c) == 2) {
                p *= b, q *= c;
            } else if(sscanf(ss, "%d", &a) == 1) {
                p *= a;
            }
            int g = gcd(p, q);
            p /= g, q /= g;
        }
        if(p == 0)
            printf("0");
        int flag = 0;
        if(p/q) printf("%d", p/q), flag = 1;
        if(p%q) {
            if(flag)    putchar('-');
            printf("%d/%d", p%q, q);
        }
        puts("");
    }
    return 0;
}