hdu 3929 Big Coefficients 解题报告 <容斥原理>

链接： http://acm.hdu.edu.cn/showproblem.php?pid=3929

Problem Description

F(x) is a polynomial in x with integer coefficients, here F(x) = (1+x)^a1 + (1+x)^a2 + ... + (1+x)^am. Given a1, a2, ... , am, find number of odd coefficients of F(x).

 

Input

The first line contains a single positive integer T( T <= 10000 ), indicates the number of test cases.
For each test case:
First line contains an integer N(1 <= N <= 15).
Second line contains N integers a1, a2, ..., am ( 0 <= ai <= 2^45 )

 

Output

For each test case: output the case number as shown and an the odd coefficients of F(x).

 

Sample Input

4 1 1 1 3 2 1 3 3 1 2 3

 

Sample Output

Case #1: 2 Case #2: 4 Case #3: 2 Case #4: 2

Hint

Case #3: (1+x) + (1+x)^3 = 2 + 4x + 3x^2 + x^3. it contains 2 odd coefficients.

Case #4: (1+x) + (1+x)^2 + (1+x)^3 = 3 + 6x + 4x^2 + x^3. it contains 2 odd coefficients.

 题意：求F( x ) 的奇系数个数

利用容斥定理 

 1 容斥原理有一般有简单的递归式
 2 
 3 dfs(int beg,set S,int sym)
 4 
 5 {
 6 
 7      ans+=num(S)*sym;
 8 
 9      for(int i=beg;i<=n;i++)
10 
11          dfs(i,S∩A[i],sym*-1);
12 
13 }
14 
15 for(int i=1;i<=n;i++)
16 
17      dfs(i,A[i],1);

代码

 1 #include <stdio.h>
 2 typedef __int64 ll;
 3 ll a[20], ans;
 4 int N;
 5 int get( ll x )
 6 {
 7     int t=0;
 8     while( x )
 9     {
10         x -= (x & -x);
11         t++ ;
12     }    
13     return t;
14 }
15 void dfs( int b, ll n, ll s )
16 {
17     ans += s*( 1<<get( n ));
18     for( int i=b+1; i<N; ++ i )    
19     {
20         dfs( i, a[i]&n, -2*s );    
21     }
22 }
23 int main( )
24 {
25     int T, t=0;
26     scanf( "%d", &T );
27     while( T-- )
28     {
29         scanf( "%d", &N );
30         for( int i=0; i<N; ++ i )
31         {
32             scanf( "%I64d", a+i );
33         }
34         ans=0;
35         for( int i=0; i<N; ++ i )
36         dfs( i, a[i], 1 );
37         printf("Case #%d: %I64d\n", ++t, ans);
38     }
39     return 0;
40 }