题意：求C (n,0),C (n,1),C (n,2)...C (n,n)中奇数的个数
分析：Lucas 定理：A、B是非负整数，p是质数。AB写成p进制：A=a[n]a[n-1]...a[0]，B=b[n]b[n-1]...b[0]。则组合数C(A,B)与C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0]) mod p同。即：Lucas (n,m,p)=C (n%p,m%p) * Lucas (n/p,m/p,p)
我是打表找规律的，就是: 2^二进制下1的个数。这定理可以求C (n, m) % p 详细解释
收获：1. 没思路时一定要打个表找找规律 2. Lucas定理
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-8-27 8:48:05
* File Name :J.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7; int main(void) {
int n;
while (scanf ("%d", &n) == 1) {
int cnt = 0;
while (n) {
if (n & 1) cnt++;
n >>= 1;
}
ll ans = 1;
for (int i=1; i<=cnt; ++i) ans *= 2;
printf ("%I64d\n", ans);
} return 0;
}