HDU 3711 Binary Number【水题】【bitset】

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本文探讨了一个算法挑战：给定两个整数集合A和B，如何为B中的每个元素找到A中与其二进制表示下不同的位数最少的元素。通过详细解析题目要求并给出实现代码，读者可以了解到解决此类问题的具体步骤。

Binary Number

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2714 Accepted Submission(s): 1573

Problem Description

For 2 non-negative integers x and y, f(x, y) is defined as the number of different bits in the binary format of x and y. For example, f(2, 3)=1,f(0, 3)=2, f(5, 10)=4. Now given 2 sets of non-negative integers A and B, for each integer b in B, you should find an integer a in A such that f(a, b) is minimized. If there are more than one such integer in set A, choose the smallest one.

Input

The first line of the input is an integer T (0 < T ≤ 100), indicating the number of test cases. The first line of each test case contains 2 positive integers m and n (0 < m, n ≤ 100), indicating the numbers of integers of the 2 sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than 1000000. The first m lines are the integers in set A and the other n lines are the integers in set B.

Output

For each test case you should output n lines, each of which contains the result for each query in a single line.

Sample Input

Sample Output

Author

CAO, Peng

Source

2010 Asia Chengdu Regional Contest

给n个数和m个数，判断对于m哪个n与它相差2进制数最少

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <bitset>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;

const int maxn = 110;
const int n = 100;

bitset<20> a[maxn], b[maxn];
int num[maxn];

int f(int p, int q)
{
	int ans = 0;
	for (int i = 0; i < 20; i++)
	{
		if (a[p][i] != b[q][i]) ans++;
	}
	return ans;
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n, m;
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; i++)
		{
			scanf("%d", num + i);
			a[i] = num[i];
		}
		for (int j = 0; j < m; j++)
		{
			int p;
			scanf("%d", &p);
			b[j] = p;
		}
		for (int i = 0; i < m; i++)
		{
			int ans = INF, val = -1;
			for (int j = 0; j < n; j++)
			{
				int temp = f(j, i);
				if (temp < ans)
				{
					ans = temp;
					val = num[j];
				}
				else if (temp == ans && (val == -1 || num[j] < val))
				{
					val = num[j];
				}
			}
			printf("%d\n", val);
		}
	}
}