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  • PAT甲级——A1082 Read Number in Chinese

    Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way. Output Fu first if it is negative. For example, -123456789 is read as Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu. Note: zero (ling) must be handled correctly according to the Chinese tradition. For example, 100800 is yi Shi Wan ling ba Bai.

    Input Specification:

    Each input file contains one test case, which gives an integer with no more than 9 digits.

    Output Specification:

    For each test case, print in a line the Chinese way of reading the number. The characters are separated by a space and there must be no extra space at the end of the line.

    Sample Input 1:

    -123456789
    

    Sample Output 1:

    Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu
    

    Sample Input 2:

    100800
    

    Sample Output 2:

    yi Shi Wan ling ba Bai

    吐槽一波,这就不能算是算法题,纯属是字符串硬处理
     1 #include <iostream>
     2 #include <string>
     3 #include <vector>
     4 using namespace std;
     5 string num[10] = { "ling","yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" };
     6 string c[6] = { "Ge","Shi", "Bai", "Qian", "Yi", "Wan" };
     7 int J[] = { 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000 };
     8 vector<string> res;
     9 int main() {
    10     int n;
    11     cin >> n;
    12     if (n == 0) {
    13         cout << "ling";
    14         return 0;
    15     }
    16     if (n < 0) {
    17         cout << "Fu ";
    18         n = -n;
    19     }
    20     int part[3];
    21     part[0] = n / 100000000;
    22     part[1] = (n % 100000000) / 10000;
    23     part[2] = n % 10000;
    24     bool zero = false; //是否在非零数字前输出合适的ling
    25     int printCnt = 0; //用于维护单词前没有空格,之后输入的单词都在前面加一个空格。
    26     for (int i = 0; i < 3; i++) {
    27         int temp = part[i]; //三个部分,每部分内部的命名规则都一样,都是X千X百X十X
    28         for (int j = 3; j >= 0; j--) {
    29             int curPos = 8 - i * 4 + j; //当前数字的位置
    30             if (curPos >= 9) continue; //最多九位数
    31             int cur = (temp / J[j]) % 10;//取出当前数字
    32             if (cur != 0) {
    33                 if (zero) {
    34                     printCnt++ == 0 ? cout << "ling" : cout << " ling";
    35                     zero = false;
    36                 }
    37                 if (j == 0)
    38                     printCnt++ == 0 ? cout << num[cur] : cout << ' ' << num[cur]; //在个位,直接输出
    39                 else
    40                     printCnt++ == 0 ? cout << num[cur] << ' ' << c[j] : cout << ' ' << num[cur] << ' ' << c[j]; //在其他位,还要输出十百千
    41             }
    42             else {
    43                 if (!zero&&j != 0 && n / J[curPos] >= 10) zero = true;   //注意100020这样的情况
    44             }
    45         }
    46         if (i != 2 && part[i] > 0) cout << ' ' << c[i + 4]; //处理完每部分之后,最后输出单位,Yi/Wan
    47     }
    48     return 0;
    49 }
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  • 原文地址:https://www.cnblogs.com/zzw1024/p/11336537.html
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