Xiao Ming ' s Hope the generalization of Lucas theorem.

　　　　　　　　　　　　　　　　　　　　　　　Xiao Ming ' s Hope

Abstract: Ask for C (n,0), C (n,1),......, C (n,n) The number of odd numbers.

Idea: Investigate C (n,m)%2.

By Lucas theorem C (A, B) =c (a[n-1],b[n-1]) *c (a[n-2],b[n-2]) * ... *c (a[0],b[0]), where A[i] is a P-binary bit

To make C (n,m)%2==1, the corresponding bit on M can be 0,1 on the 1 bit on N, (c (1,0) =c () =1).

For bits 0 on n, the corresponding bit on M must be 0. (c (0,0) =1,c (0,1) =0).

So the title of the ans=2^cnt (CNT is n of the number of bits in 1).

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6 using namespacestd;7typedefLong LongLL;8 9 intMain ()Ten { One     intN; A      while(SCANF ("%d", &n)! =EOF) -     { -         intCnt=0; the          while(n) -         { -             if(n&1) -cnt++; +N/=2; -         } +printf"%d\n",1<<CNT); A     } at     return 0; -}

binomial coefficients

Topic abstraction: Judging the value of C (n,m)%2.

Idea: Lucas theorem.   With M as the research object, for the 0 bits on the binary of M, there is no effect on singularity. Examine the 1 bits on the binary of M. For the result to be 1, the corresponding bit on n must be 1.

Ans= (n&m) ==m?1:0;

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6 using namespacestd;7typedefLong LongLL;8 9 intMain ()Ten { One     intn,m; A      while(SCANF ("%d%d", &n,&m)! =EOF) -     { -         if((n&m)! =m) theprintf"0\n"); -         Else -printf"1\n"); -     } +     return 0; -}

Xiao Ming ' s Hope the generalization of Lucas theorem.