Day 0x19 UVa10929 You can say 11

• Virtual Judge
• ZeroJudge

题意

• 输入一整数，判断是否为 11 的倍数
• 需要注意的有：
  1. 重复输入一整数 N 直到 0
  2. N 最大可到 1000 位
  3. 输出格式
     • N is a multiple of 11.
     • N is not a multiple of 11.

解法

• 如何快速判断一个数是不是 11 的倍数？
• 重复输入一字串 n (因为整数存不下) 直到 0

  char n[1000] = {0};
  
  while(scanf("%s", n)){
      if(*n == '0' && strlen(n) == 1){
          break;
      }
      else{
          ...
      }
  }
  

• 透过 for 回圈及回圈变数 i 控制一加一减每个位数，要记得是字元所以要 - '0'

  int i;
  int difference = 0;
  
  for(i = 0; i < strlen(n); i++){
      if(i % 2 == 0){
          difference = difference + (n[i] - '0');
      }
      else{
          difference = difference - (n[i] - '0');
      }
  }
  

• 最後判断是否为 11 的倍数再输出结果

  if(difference % 11 == 0){
      flag = true;
  }
  
  if(flag){
      printf("%s is a multiple of 11.\n", n);
  }
  else{
      printf("%s is not a multiple of 11.\n", n);
  }
  

• C code

  #include<stdio.h>
  #include<string.h>
  #include<stdbool.h>
  
  int main(){
  
      char n[1000] = {0};
  
      while(scanf("%s", n)){
          if(*n == '0' && strlen(n) == 1){
              break;
          }
          else{
  
              int i;
              bool flag = false;
              int difference = 0;
  
              for(i = 0; i < strlen(n); i++){
                  if(i % 2 == 0){
                      difference = difference + (n[i] - '0');
                  }
                  else{
                      difference = difference - (n[i] - '0');
                  }
              }
  
              if(difference % 11 == 0){
                  flag = true;
              }
  
              if(flag){
                  printf("%s is a multiple of 11.\n", n);
              }
              else{
                  printf("%s is not a multiple of 11.\n", n);
              }
          }
      }
  
      return 0;
  }