1 solutions

  • 0
    @ 2025-11-5 19:23:34

    C :

    #include <stdio.h>
    int main()
    {
    int i;
    float m=0,n=0,sum=0;
    scanf("%f %f",&m,&n);
    for(i=1;i<=n;i++)
       {
       sum+=(m+m/2);
       m=m/2;
       }
        printf("%.2f %.2f\n",m,sum-m);
    return 0;
    }
    

    C++ :

    #include<stdio.h>
    int main()
    {
        float sum,m;
        int n;
        scanf("%f%d",&m,&n);
        sum=m;
        while(n--)
            {
                m=m/2;
                sum+=2*m;
            }
        printf("%.2f %.2f\n",m,sum-2*m);
    }
    
    
    • 1

    C语言程序设计教程(第三版)课后习题6.9

    Information

    ID
    19328
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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