hdu2608 0 or 1

/*分析:假设数n=2^k*p1^s1*p2^s2*p3^s3*...*pi^si;//k,s1...si>=0,p1..pi为n的素因子所以T[n]=(2^0+2^1+...+2^k)*(p1^0+p1^1+...+p1^s1)*...*(pi^0+pi^1+...+pi^si);显然(2^0+2^1+...+2^k)%2=1,所以T[n]是0或1就取决于(p1^0+p1^1+...+p1^s1)*...*(pi^0+pi^1+...+pi^si)而p1...pi都是奇数(除2之外的素数一定是奇数),所以(pi^0+pi^1+...+pi^si)只要有一个si为奇数(i=1...i)则(pi^0+pi^1+...+pi^si)%2=0,则T[n]%2=0//若si为奇数,则pi^si+1为偶数,pi^1+pi^2+...+pi^(si-1)为偶数(偶数个奇数和为偶数)所以要T[n]%2=1,则所有的si为偶数，则n=2^(k%2)*m^2;//m=2^(k/2)*p1^(s1/2)*p2^(s2/2)*...*pi^(si/2)所以只要n为某个数的平方或者某个数的平方和则T[n]%2=1,只要统计n的个数即可另外对T[n]进行打表能找到规律也行*/#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<cmath>#include<iomanip>#define INF 99999999using namespace std;const int MAX=10;int main(){    int t, n;    scanf("%d", &t);    while ( t-- )    {        scanf("%d", &n);        int sum=(int)sqrt(n*1.0)+(int)sqrt(n*1.0/2);        //cout<<sum%2<<endl;        printf("%d\n", sum%2);    }    return 0;}