解题思路:
注意事项:
参考代码:
#include<stdio.h> char dic[10][10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"}; int open=0; void readYi(int n) { //传进来的最多有2位数 // n%10;//1位(从右往左数) // n/10%10; if(n>9) { if(n/10%10==1) printf("shi %s yi ",dic[n%10]); else printf("%s shi %s yi ",dic[n/10%10],dic[n%10]); } else { //n<=9 printf("%s yi ",dic[n]); } } void readQian(int n) { if(n/1000%10!=0) { printf("%s qian ",dic[n/1000%10]); open=1; } if(n/100%10!=0) { printf("%s bai ",dic[n/100%10]); open=1; } if(n/10%10!=0&&n/100%10!=0) { printf("%s shi ",dic[n/10%10]); open=1; } else if(n/10%10!=0&&n/100%10==0&&open==1) { printf("ling %s shi ",dic[n/10%10]); open=1; } else if(n/10%10!=0&&n/100%10==0&&open==0&&n/10%10==1) { printf("shi ",dic[n/10%10]); open=1; } else if(n/10%10!=0&&n/100%10==0&&open==0) { printf("%s shi ",dic[n/10%10]); open=1; } if(n%10!=0&&(n/10%10!=0)) { printf("%s ",dic[n%10]); open=1; } else if(n%10!=0&&(n/10%10==0)&&open==1) { printf("ling %s ",dic[n%10]); open=1; } else if(n%10!=0&&(n/10%10==0)&&open==0) { printf("%s ",dic[n%10]); open=1; } } void readQianwan(int n) { //n%10;//第1位(从右往左数) // n/10%10; // n/100%10; // n/1000%10; readQian(n); printf("wan "); } int main() { int n; scanf("%d",&n); if(n/100000000!=0) readYi(n/100000000); if(n/10000%10000!=0) readQianwan(n/10000%10000); readQian(n%10000); return 0; }
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