思路:

用字符數組來存儲這個大整數

由它的範圍可以知道每一項相加都不會大於int的最大值,所以可以用int存儲最後各位相加的值

將sum值輸入到一個字符數組,然後用另一個數組映射對應的字符串即可

#include <cstdio>

using namespace std;


int main() {
  char num[111]={‘\0‘};
  scanf("%s",num);
  int sum=0;

  for(int i=0;num[i]!=‘\0‘;i++) {
    sum+=(num[i]-‘0‘);
  }
  
 
//printf("%d\n",sum);
  const char* res[]={"zero","one","two","three","four","five","six","seven","eight","nine"};
  
  char tem[10]={‘\0‘};
  sprintf(tem,"%d",sum);
  //printf("%s\n",tem);
  for(int i=0;tem[i]!=‘\0‘;i++) {
    tem[i+1]==‘\0‘? printf("%s",res[tem[i]-‘0‘]):printf("%s ",res[tem[i]-‘0‘]);
  }
  printf(
 
"\n");
  return 0;
}

1005. Spell It Right (20)