3 条题解

  • 0
    @ 2023-12-7 17:42:08

    #include<stdio.h> #include<string.h> int main() { int n; char s[1000005]; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%s", s); int la = strlen(s); for (int i = 0; i < la; i++) { s[i] = s[i] - '0'; }//将每一个字符变成数值 for (int i = 0; i < la-1; i++) { if (s[i] == 9) { s[i] = 0; } if (s[i] < 9) { s[i + 1] = (10 * s[i] + s[i + 1]) % 9; s[i]=0; }//每进进一位,取余一次 } printf("%d\n",s[la-1]); }

    return 0;
    

    }

    • 0
      @ 2023-10-12 21:31:37
      #include<stdio.h>
      #include<string.h>
      int main(){
      	int n;
      	scanf("%d",&n);
      	while(n--){
      		char nstr[1000005];
      		int num = 0;
      		scanf("%s", nstr); 
      		int len = strlen(nstr);
      		for (int i = 0; i < len; i++)
      		num = (num * 10 + nstr[i] - '0') % 9;
      /*从高位到低位取余,
      逐位计算各个位数之和及其与9的余数*/
      		printf("%d\n", num%9);
      		memset(nstr, 0, sizeof(nstr));
      	}
      	return 0;
      }
      
      • 0
        @ 2023-10-9 21:28:59
        #include <stdio.h>
        
        char number[1000005];
        
        int main()
        {
            int i, n, testNum;
        
            scanf("%d\n", &testNum);
            while (testNum-- != 0)
            {
                scanf("%s", number);
                for (i = n = 0; number[i] != '\0'; i++)
                {
                    n += number[i] - '0';
                }
                printf("%d\n", n%9);
            }
            return 0;
        }
        
        • 1

        信息

        ID
        166
        时间
        3000ms
        内存
        128MiB
        难度
        7
        标签
        (无)
        递交数
        945
        已通过
        212
        上传者