蓝桥杯_基础训练_数的读法

有点恶心的一道题，做了整整半天才完全对。

思路：

把输入的数从右往左每4位进行分割。然后判断单位级数，输出即可。

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#include <iostream> #include <cstring> #include <algorithm>   using namespace std;   int main() {     string a;     string num[]={"ling ","yi ","er ","san ","si ","wu ","liu ","qi ","ba ","jiu "};    //存数的拼音     string wei[]={"","shi ","bai ","qian "};              //单位读法     string kk[]={"","wan ","yi "};                //级数读法 万 亿     string b[3];      //存分割之后的数，从右到左每四位分为一个字符串     while(cin>>a)     {         int alen=a.length();         for(int i=0;i<3;i++)             b[i].clear();         int blen=0;         int k=1;         for(int i=a.length()-1;i>=0;i--)    //每四位进行分割字符         {             if(k<5)             {                     b[blen]+=a[i];                     k++;             }             else             {                 blen++;                 k=1;                 i++;             }         }         for(int i=blen;i>=0;i--)         {             reverse(b[i].begin(),b[i].end());   //翻转分割之后的每组字符串             //cout<<b[i]<<endl;         }         for(int j=blen;j>=0;j--)    //下标小的级数高，比如123456的分割结果：b[0]="12"，b[1]="3456"；         {             for(int i=0;i<b[j].length();i++)    //循环输出每组字符串             {                 if(i==b[j].length()-1&&b[j][i]-'0'==0)   //如果到最后一位并且该位为0,                 {                     continue;                 }                 else                 {                     if(b[j][i]-'0'==0&&b[j][i+1]-'0'==0)   //如果中间有多个0                         continue;                     if(b[j][i]-'0'==0)                   //如果该位为0，只输出数                     {                         cout<<num[b[j][i]-'0'];                         continue;                     }                     else                     {                         if(b[j].length()==2&&i==0&&b[j][i]=='1')   //如果当前字符串的长度为2，并且最高位为1。处理十几的读法，比如12读 shi er，不读yi shi er。                         {                             cout<<wei[b[j].length()-i-1];                         }                         else                             cout<<num[b[j][i]-'0']<<wei[b[j].length()-i-1];                     }                 }             }             cout<<kk[j];         }         cout<<endl;     }     return 0; }