Hdu4349--xiao Ming ' s Hope (number theory)

Enter an n (1<=n<=108), C (n,0), C (n,1), C (n,2) ... C (N,n) has the number of odd numbers.

Lacus theorem http://blog.csdn.net/acm_cxlove/article/details/7844973

A, B is a non-negative integer and P is a prime number. AB is written in P-system: a=a[n]a[n-1]...a[0],b=b[n]b[n-1]...b[0].

The combined number C (A, B) and C (A[n],b[n]) *c (a[n-1],b[n-1]) *...*c (a[0],b[0]) MODP the same

So in this topic C (n, x) will n and X into 2 C (0,0) =1,c (0,1) =0,c (1,0) =1,c (a) =1

So if C (n,x) is 1, then n is 0 where x must be 0,n 1, where x is 0 or 1

The answer is 2^ (the binary of n represents the number of 1)

#include <iostream> #include <cstdio> #include <cmath>using namespace Std;int main () {    int n, c;    while (~SCANF ("%d", &n)) {        c = 0;        while (n) {            if (N & 1) ++c;            n >>= 1;        }        printf ("%d\n", (int) POW (2, c));    }    return 0;}

　　

Hdu4349--xiao Ming ' s Hope (number theory)