1 solutions

  • 0
    @ 2025-11-5 20:14:04

    C++ :

    //384K	0MS
    #include<stdio.h>
    #include<math.h>
    int gcd(int a,int b)
    {
        if(!b)return a;
        return gcd(b,a%b);
    }
    int main()
    {
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
        int c,s;
        while(scanf("%d%d",&c,&s)!=EOF)
        {
            int sum=0;
            for(int i=1;i<=s;i++)
            {
                int tmp=gcd(s,i);//第i次旋转的循环节数
                sum+=(int)(pow(c*1.0,tmp*1.0));
            }
            if(s&1)sum+=(int)(s*pow(c*1.0,(s+1)/2.0));//s为奇数,共有s个循环节数均为(s+1)/2的置换
            else//当s为偶数
            {
                sum+=(int)((s/2)*pow(c*1.0,(s+2)/2.0));//第一种循环节数均为(s+2)/2
                sum+=(int)((s/2)*pow(c*1.0,s/2.0));//第二种循环节数均为s/2
            }
            sum/=(2*s);
            printf("%d\n",sum);
        }
        return 0;
    }
    
    
    • 1

    Information

    ID
    20163
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    (None)
    Tags
    # Submissions
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