9 条题解

  • 0
    @ 2025-12-10 21:11:28
    #include <stdio.h>
    int main()
    {
        int n, x, y;
        scanf("%d %d %d", &n, &x, &y);
            int num;
        num = n -(y+x-1)/x;
        if(num<0){
            num=0;
        }
        printf("%d", num);
        return 0;
    }
    
    • 0
      @ 2025-9-23 16:07:25
      #include<stdio.h>
      int main()
      {
          int n,x,y;
          scanf("%d%d%d",&n,&x,&y);
          printf("%d",n-((y-1)/x+1));
      }
      
      • 0
        @ 2024-10-22 19:15:12

        这个为什么不可以? #include<stdio.h> int main() { int n,x,y; scanf("%d %d %d",&n,&x,&y); (int)n=n-y*1.0/x; printf("%d",n); return 0;

        }

        • 0
          @ 2024-4-24 22:27:56
          #include <stdio.h>
          int main()
          {
              
              float b,c;
              int a;
              scanf("%d%f%f",&a,&b,&c);
              int n=a-c/b;
              printf("%d",n);
              return 0;
          }//
          
          • 0
            @ 2023-9-16 23:18:04

            #include<stdio.h> int main() { int n,x,y,sum; scanf("%d %d %d",&n,&x,&y); if(y%x==0) sum=n-y/x; else {sum=n-y/x-1; } printf("%d",sum); return 0; }

            • 0
              @ 2023-8-29 22:58:27
              #include <stdio.h> 
              int main()
              {
                  float x,y,n,c;
                  scanf("%f%f%f",&n,&x,&y);
                  c=y/x;
                  n=n-c;
                  printf("%d",(int)n);
                  return 0;
              }
              
              • 0
                @ 2023-8-28 15:32:56

                #include <stdio.h>

                int main() { int a = 0; int b = 0; int c = 0; scanf("%d %d %d", &a, &b, &c); int x; x = a - (1.0 /b)* c; printf("%d", x); return 0; }//看了其他人的题解我感觉我这个有点问题————-----------

                • 0
                  @ 2023-8-13 21:42:35
                  #include<stdio.h>
                  int main(){
                      int n,x,y;
                      scanf("%d %d %d",&n,&x,&y);
                      if(y % x == 0){
                         printf("%d", n - (y / x));
                      } else {
                          printf("%d",n - ((y / x) + 1));
                      }
                      return 0;
                  }
                  
                  • 0
                    @ 2023-8-1 22:05:17
                    #include<stdio.h>
                    int main(){
                        int n,x,y;
                        scanf("%d %d %d",&n,&x,&y);
                        if(y % x == 0){
                           printf("%d", n - (y / x));
                        } else {
                            printf("%d",n - ((y / x) + 1));
                        }
                        return 0;
                    }
                    
                    • 1

                    信息

                    ID
                    45
                    时间
                    1000ms
                    内存
                    256MiB
                    难度
                    4
                    标签
                    递交数
                    3293
                    已通过
                    1442
                    上传者