8 条题解

  • 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
                  2934
                  已通过
                  1314
                  上传者