Vijos 1696题：数与连分数

描述

写一个程序…可以实现在连分数和分数之间的互相转换…

样例1

样例输入1

[2;3,7]
51/22

样例输出1

51/22
[2;3,7]

提示

多组测试数据:
…每一个测试点有多组数据…数据的组数不超过100组…
约分:
计算结果最后是要约分的…但是..在分数转向连分数的时候..
我们不保证输入的数据是约分…..
连分数的输入格式:
连分数的输入格式是按照wiki里写的…
第一个位置的分号应该是为了避免可能的歧义~..
Range:
数字的规模都很小…
也就是它们都不会超过longint范围里….
包括中间的数据…
连分数的项数也不会超过100项..

---

#include <iostream>#include <string>#include <memory.h>using namespace std;string s;int main(){    std::ios::sync_with_stdio(false);    while (cin >> s)    {        int a[100], t = 0;        bool flag = false;        memset(a, 0, sizeof(a));        if (s[0] == '[')        {            for (int i = 1; i < s.size(); i++)            {                if (s[i] >= '0'&&s[i] <= '9')                {                    flag = true;                    a[t] = a[t] * 10 + (int)(s[i] - '0');                }                else                {                    t++;                    flag = false;                }            }            int x = a[t], y = 1;            for (int i = t; i >= 0; i--)            {                int temp = x;                x = y + a[i] * x;                y = temp;            }            int max = x, min = y;            while (max%min != 0)            {                int temp = min;                min = max%min;                max = temp;            }            x = x / min, y = y / min;            if (y == 1)                cout << x << endl;            else                cout << x << "/" << y << endl;        }        else        {            for (int i = 0; i < s.size(); i++)            {                if (s[i] >= '0'&&s[i] <= '9')                {                    flag = true;                    a[t] = a[t] * 10 + (int)(s[i] - '0');                }                else                    t++;            }            if (t == 0)                cout << '[' << a[0] << ']' << endl;            else            {                int max = a[0], min = a[1];                while (max%min != 0)                {                    int temp = min;                    min = max%min;                    max = temp;                }                int x = a[0] / min, y = a[1] / min;                t = 0;                cout << "[";                while (x != 1)                {                    int div = x / y, rem = x%y;                    if (t == 0)                        cout << div;                    else                        if (t == 1)                            cout << ";" << div;                        else                            cout << "," << div;                    x = rem;                    if (x != 1)                    {                        int temp = y;                        y = x;                        x = temp;                    }                    t++;                }                if (y != 0)                    cout << ',' << y;                cout << ']' << endl;            }        }    }    return 0;}