2012 Multi-University Training Contest 5-1011 hdu4349
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题目大意:给一个数N(10^8),求 C(n,0),C(n,1),C(n,2)...C(n,n).中有多少个奇数。
解体思路:本题为Lucas定理推导题,我们分析一下 C(n,m)%2,那么由lucas定理,我们可以写
成二进制的形式观察,比如 n=1001101,m是从000000到1001101的枚举,我们知道在该定理中
C(0,1)=0,因此如果n=1001101的0对应位置的m二进制位为1那么C(n,m) % 2==0,因此m对应n为0的
位置只能填0,而1的位置填0,填1都是1(C(1,0)=C(1,1)=1),不影响结果为奇数,并且保证不会
出n的范围,因此所有的情况即是n中1位置对应m位置0,1的枚举,那么结果很明显就是:2^(n中1的个数)
代码:
#include <iostream>#include <cstdio>using namespace std;int main(){ int n,res; while(scanf("%d",&n)!=EOF) { res = 1; while(n) { if(n%2) { res*=2; } n /= 2; } printf("%d\n",res); }}
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