ZOJ 3870 Team Formation

题意：给出n个数，在其中任意挑出两个数进行异或，判断有多少种情况使得异或的值大于原数。

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5518

思路：转化为二进制，对位进行拆分，求出每一位的0的个数a以及二进制位数b，输出sum(a*b)

注意点：无

以下为AC代码：

Run ID	Submit Time	Judge Status	Problem ID	Language	Run Time(ms)	Run Memory(KB)	User Name
3945006	2015-04-26 18:38:13	Accepted	3870	C++0x	380	272	luminus

/*
***********************************************
*# @Author  : Luminous11 (573728051@qq.com)
*# @Date    : 2015-04-26 18:25:24
*# @Link    : http://blog.csdn.net/luminous11 ***********************************************
*/

#include <bits/stdc++.h>
#define clr(a, v) memset( a , v , sizeof(a) )
using namespace std;
const double eps = 1e-10;
const double pi = acos(-1.0);
int num0[35];
int num1[35];
void calc ( int tmp )
{
int p = 1;
while ( tmp ){
if ( ( tmp & 1 ) == 0 ){
num0[p] ++;
}
tmp >>= 1;
p ++;
}
num1[p-1] ++;
}
int main()
{
//ios::sync_with_stdio ( false );
int T;
scanf ( "%d", &T );
while ( T -- ){
int n;
scanf ( "%d", &n );
memset ( num0, 0, sizeof ( num0 ) );
memset ( num1, 0, sizeof ( num1 ) );
int tmp;
for ( int i = 0; i < n; i ++ ){
scanf ( "%d", &tmp );
calc ( tmp );
}
long long ans = 0;
for ( int i = 0; i < 35; i ++ ){
ans += (long long)num0[i] * (long long)num1[i];
}
printf ( "%lld\n", ans );
}
return 0;
}