又做了几道HOJ水题。。。

1. 1010: 纯粹的数学题，就在输出那要记住一点，C++中要固定输出浮点数的位数，需要使用iomanip库，使用setprecision()函数即可。

   代码如下：

    #include <iostream>
    #include <iomanip>
    using namespace std;
   
    int main(){
        int hou, min;
        double min_ang, hou_ang, angle;
        while(cin >> hou >> min){
            if(hou == 0 && min == 0)
                break;
            else{
                min_ang = (double)(min * 6);
                hou_ang = (double)(hou % 12 * 30) + ((double)min) / 60.0 * 30.0;
                if(hou_ang >= min_ang)
                    angle = hou_ang - min_ang <= 180 ? hou_ang - min_ang : 360 + min_ang - hou_ang;
                else
                    angle = min_ang - hou_ang <= 180 ? min_ang - hou_ang : 360 - min_ang + hou_ang;
                cout << "At " << hou << ':';
                if(min < 10)
                    cout << 0 << min;
                else
                    cout << min;
                cout << " the angle is " << fixed << setprecision(1) << angle << " degrees." << endl;
            }
        }
    }  
   

2. 1078: setprecision()固定输出精度后，后面的并不是简单舍去，而是四舍五入。

    #include <iostream>
    #include <iomanip>
    using namespace std;
   
    int main(){
        int tmp;
        double result;
        cout << "n e\n- -----------\n0 1\n1 2\n2 2.5" << endl;
        for(int i = 3; i < 10; i++){
            cout << i << ' ';
            tmp = 2;
            result = 2.5;
            for(int j = 3; j <= i; j++){
                tmp *= j;
                result += 1.0 / ((double)tmp);
            }
            cout << fixed << setprecision(9) << result << endl;
        }
    }  
   

3. 1034: 这是一道重点在于字符串的题目，正好复习一下C/C++的字符串相关的函数。总体来说不难，只要肯花时间就行，我已经能够按顺序输出数字了，但是后面的输出格式话实在太烦了，不想弄了，今天就到此为止吧。

    #include <iostream>
    #include <cstring>
    #include <string>
    #include <iomanip>
    using namespace std;
   
    void print(char d[], int s, int l){
        char front[20], rear[20];
        int tmp, front_num = 0, rear_num = 0, results[20], result;
   
        for(int i = 0; i < s; i++){
            front[i] = d[i];
            tmp = (int)front[i] - 48;
            for(int j = 1; j < s - i; j++)
                tmp *= 10;
            front_num += tmp;
        }
        front[s] = '\0';
   
        for(int i = s + 1; i < l; i++){
            rear[i - s - 1] = d[i];
            tmp = (int)d[i] - 48;
            for(int j = 1; j < l - i; j++)
                tmp *= 10;
            rear_num += tmp;
        }
        rear[l - s] = '\0';
   
        switch(d[s]){
            case '+':
                result = front_num + rear_num;
                if(s >= l - s){
                    cout << front_num << endl;
                    for(int i = 0; i < s - l + s; i++)
                        cout << ' ';
                    cout << '+' << rear_num << endl;
                    for(int i = 0; i < s; i++)
                        cout << '-';
                    cout << '\n' << setw(s) << result << endl;
                }else{
                    for(int i = 0; i < l - s - s; i++)
                        cout << ' ';
                    cout << front_num << endl;
                    cout << '+' << rear_num << endl;
                    for(int i = 0; i < l - s; i++)
                        cout << '-';
                    cout << '\n' << setw(l - s)  << result << endl;
                }
                break;
        case '-':
            result = front_num - rear_num;
            cout << front_num << endl;
            cout << '-' << rear_num << endl;
            for(int i = 0; i < strlen(rear); i++)
                cout << '-';
            cout << '\n' << result << endl;
            break;
        case '*':
            result = front_num * rear_num;
            cout << front_num << endl;
            cout << '*' << rear_num << endl;
            for(int i = 0; i < strlen(rear); i++)
                cout << '-';
            cout << endl;
            for(int i = l - s - 2; i >= 0; i--)
                cout << front_num * ((int)rear[i] - 48) << endl;
            for(int i = 0; i < strlen(rear); i++)
                cout << '-';
            cout << '\n' << result << endl;
        break;
        }
    }
   
    int main(){
        int count, sym, length;
        cin >> count;
        char input[40];
        while(count--){
            input[0] = '\0';
            cin >> input;
            length = strlen(input);
            for(int i = 0; i < length; i++){
                if((int)input[i] < 48 || (int)input[i] > 57)
                sym = i;
            }
            print(input, sym, length);
        }
    }