首页 > 代码库 > SRM 621 D2L3: MixingColors, math

SRM 621 D2L3: MixingColors, math

题目:http://community.topcoder.com/stat?

c=problem_statement&pm=10409&rd=15854


利用高斯消元求线性空间的基,也就是求矩阵的秩。

代码:

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
using namespace std;

#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair
#define FOREACH(it, X) for(__typeof((X).begin()) it = (X).begin(); it != (X).end(); ++it)

/*************** Program Begin **********************/

class MixingColors {
public:
	int minColors(vector <int> colors) {
		int res = 0;
		int n = colors.size();
		bool used[55];
		memset(used, 0, sizeof(used));

		for (int i = 31; i >= 0; i--) {
			int pivot = 0;
			for (int j = 0; j < n; j++) {
				if (used[j]) {
					continue;
				}
				if ((colors[j] >> i) & 0x1) {
					if (!pivot) {
						pivot = colors[j];
						used[j] = true;
						++res;
					} else {
						colors[j] ^= pivot;
					}
				}
			}
		}

		return res;
	}
};

/************** Program End ************************/


SRM 621 D2L3: MixingColors, math