11 条题解

  • 1
    @ 2023-8-18 22:49:53
    #include<stdio.h>
    int main()
    {
    int a=0,b=0,m,n;
    scanf("%d %d",&n,&m);
    b=(m-2*n)/2;
    a=n-b;
    printf("%d %d",a,b);
    return 0;
    }
    
    • 1
      @ 2022-9-22 20:46:53

      #include<stdio.h> int main() { int n,m,a,b; scanf("%d%d",&n,&m); a=(4*n-m)/2; b=n-a; printf("%d %d",a,b); return 0; }

      • 0
        @ 2024-8-20 1:23:38

        #include <stdio.h> int main(){ int n,m,a,b; scanf("%d %d",&n,&m); a=2*n-m/2; b=m/2-n; printf("%d %d",a,b); return 0; }

        • 0
          @ 2024-4-24 22:06:15
          #include <stdio.h>
          int main()
          {
              int a,b,m,n;
              scanf("%d%d",&a,&b);
              m=(4*a-b)/2;
              n=a-m;
              printf("%d %d",m,n);
              return 0;
          }//
          
          • 0
            @ 2024-1-9 15:13:07

            ##设鸡有x只,则兔有(n-x)只 ##2x+(n-x)4=m ##2x+4n-4x=m ##4n-m=2x ##x=(4n-m)/2 n,m=map(int,input().split()) print(int((4n-m)/2),int(n-(4n-m)/2))

            • 0
              @ 2023-9-4 22:01:21

              int main() {

              int jttl1(int n,int m);
              int jttl2(int n,int m);
              int n,m,a,b;
              scanf("%d %d",&n,&m);
              a=jttl1(n,m);
              b=jttl2(n,m);
              printf("%d %d",a,b);
              return 0;
              

              }

              int jttl1(int x,int y)

              { return 2*x-0.5**y; }

              int jttl2(int x,int y)

              { return 0.5*y-x; }

              • 0
                @ 2023-8-16 13:34:25

                #include <stdio.h>

                int main() { int n,m,a,b;//a是鸡的数量,b是兔子的数量 scanf("%d %d",&n, &m);//n个头,m个脚 b=(m-2*n)/2; a=n-b;

                printf("%d %d",a, b);
                
                return 0;
                

                }

                • 0
                  @ 2023-8-15 17:53:17

                  #define _CRT_SECURE_NO_DEPRECATE 1 #include<stdio.h> int main() { int n, m, a, b; scanf("%d%d",&n,&m); b = (m - 2*n) / 2; a = n - b; printf("%d %d",a,b); return 0; }

                  • 0
                    @ 2023-8-15 11:14:44

                    #include<stdio.h> int main (){ int n,m; scanf("%d %d",&n,&m); int a,b; b=(m-2*n)/2; a=n-b; printf("%d %d",a,b);

                    return 0; }

                    • 0
                      @ 2023-8-8 9:17:54

                      #include<stdio.h>** ** int main() {** int n,m,a,b;** ** scanf("%d %d",&n,&m);** ** a=(4*n-m)/2;** ** b=n-a;** printf("%d %d",a,b); ** return 0;** }

                      • 0
                        @ 2023-1-1 15:17:03

                        #include<stdio.h> int main() { int n,m,a,b; scanf("%d %d",&n,&m); for(a=0;a<=n;a++) { for(b=n-a;b>=0;b--) { if(a+bn && a2+b4m) printf("%d %d",a,b); break; } } }

                        • 1

                        信息

                        ID
                        19
                        时间
                        1000ms
                        内存
                        128MiB
                        难度
                        5
                        标签
                        递交数
                        6534
                        已通过
                        2344
                        上传者