3 solutions

  • 0
    @ 2023-10-12 14:38:46

    #include<bits/stdc++.h>

    using namespace std;

    int x,y,n;

    int main() {

    cin>>x>>y;

    for(int m=1;m<=x;m++){

    int n=x-m;

    if((2m+4n)==y) cout<<m<<' '<<n; }

    return 0;

    } /* 1.鸡兔共x只,设鸡m只,则兔 为(x-m)只,则:

    2m+4(x-m)=y

    */

    • 0
      @ 2023-10-12 14:38:00

      数学题

      • 0
        @ 2023-10-12 14:37:43

        #include<bits/stdc++.h> using namespace std; int x,y,n; int main() { cin>>x>>y; for(int m=1;m<=x;m++){ int n=x-m; if((2m+4n)==y) cout<<m<<' '<<n; } return 0; } /* 1.鸡兔共x只,设鸡m只,则兔为(x-m)只,则:

        2m+4(x-m)=y
        

        */

        • 1

        Information

        ID
        430
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        8
        Tags
        # Submissions
        14
        Accepted
        6
        Uploaded By