ZOJ 3870 Team Formation

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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 IDSubmit TimeJudge StatusProblem IDLanguageRun Time(ms)Run Memory(KB)User Name39450062015-04-26 18:38:13Accepted 3870C++0x380272luminus

/* ************************************************# @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;}

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