LightOJ 1336Sigma Function

Posted 2020-09-03 BBBob

/*
 LightOJ 1336    Sigma Function
 http://lightoj.com/login_main.php?url=volume_showproblem.php?problem=1336
数论 奇偶性
题目求f(n)为偶数的个数
我们发现如果f(n)为奇数，则n为x^2,2*x^2,2^x三种形式，
因为2^x中已经包含剩下两种，
所以只需求x^2和2*x^2的个数即可求得答案。
 *
 打了半天表发现没用233333
 *
 */


#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
    //freopen("a.in","r",stdin);
    int t;
    scanf("%d",&t);
    long long n;
    int j=1;
    while(t--)
    {
        scanf("%lld",&n);
        long long ans=0;
        for(long long i=1;i*i<=n;i++)
        {
            if(2*i*i<=n)
                ans++;
            ans++;
        }       
        ans=n-ans;
        printf("Case %d: %lld\n",j,ans); 
        j++;
    }
    return 0;

}





//打表
/*#include <cstdio>*/
//#include <algorithm>
//using namespace std;
//const int Nmax=10001;
//int prime[Nmax];
//int is_prime[Nmax];
//int cnt;
//int m[Nmax];
//int times[Nmax];
//int num;
//void get_prime()
//{
    //for(int i=2;i<Nmax;i++)
        //is_prime[i]=1;
    //for(int i=2;i<Nmax;i++)
    //{
        //if(is_prime[i])
        //{
            //prime[++cnt]=i;
            //for(int j=2;j*i<Nmax;j++)
                //is_prime[i*j]=0;
        //}
    //}
//}

//int pow(int base,int n)
//{
    //int ans=1;
    //while(n>0)
    //{
        //if(n&1)
            //ans*=base;
        //base*=base;
        //n>>=1;
    //}
    //return ans;
//}

//int check()
//{
    //for(int i=1;i<=num;i++)
    //{
        //int a=pow(m[i],times[i]+1)-1;
        //int b=m[i]-1;
        //a/=b;
        //if(!(a&1))
            //return 1;
    //}
    //return 0;
//}

//void get(int n)
//{
    //num=0;
    //int i=1;
    //while(prime[i]*prime[i]<=n)
    //{
        //if(n%prime[i]==0)
        //{
            //++num;
            //times[num]=0;
            //m[num]=prime[i];
        //}
        //while(n%prime[i]==0)
        //{
            //n/=prime[i];
            //times[num]++;
        //}
        //i++;
    //}
    //if(n!=1)
    //{
        //++num;
        //times[num]=1;
        //m[num]=n;
    //}
//}


//void print(int n)
//{ 
    //int ans=0;
    //for(int i=2;i<=n;i++)
    //{
        //get(i);
        //ans+=check();
    //} 
    //printf("%d:%d\n",n,ans);
//}

//int main()
//{   
    //get_prime();
    //for(int n=1;n<=1000;n++)
    //{
        //print(n);
    //}
    //return 0;
/*}*/