USACO hamming

考试周终于过去了一半，可以继续写USACO了。

先来看一下题目吧。

Hamming Codes
Rob Kolstad

Given N, B, and D: Find a set of N codewords (1 <= N <= 64), each of length B bits (1 <= B <= 8), such that each of the codewords is at least Hamming distance of D (1 <= D <= 7) away from each of the other codewords. The Hamming distance between a pair of codewords is the number of binary bits that differ in their binary notation. Consider the two codewords 0x554 and 0x234 and their differences (0x554 means the hexadecimal number with hex digits 5, 5, and 4):

        0x554 = 0101 0101 01000x234 = 0010 0011 0100
Bit differences: xxx  xx

Since five bits were different, the Hamming distance is 5.

PROGRAM NAME: hamming

INPUT FORMAT

N, B, D on a single line

SAMPLE INPUT (file hamming.in)

16 7 3

OUTPUT FORMAT

N codewords, sorted, in decimal, ten per line. In the case of multiple solutions, your program should output the solution which, if interpreted as a base 2^B integer, would have the least value.

SAMPLE OUTPUT (file hamming.out)

0 7 25 30 42 45 51 52 75 76
82 85 97 102 120 127

这条题目不难，由于数据量很小，因此使用的方法是直接遍历。

/**
ID: njuwz151
TASK: hamming
LANG: C++
*/
#include <iostream>
#include <cstdio>using namespace std;const int maxn = 9;int count(int a, int b);int main() {freopen("hamming.in", "r", stdin);freopen("hamming.out", "w", stdout);int n, b, d;cin >> n >> b >> d;int result[1 << maxn];result[0] = 0;int n_ptr = 1;for(int i = 1; i < (1 << b); i++) {if(n_ptr > n) {break;}bool can_add = true;for(int j = 0; j < n_ptr; j++) {if(count(i, result[j]) < d) {can_add = false;break;}}if(can_add) {result[n_ptr] = i;n_ptr++; }}for(int i = 0; i < n - 1; i++) {cout << result[i];if((i + 1) % 10) {cout << " ";} else {cout << endl;}}cout << result[n - 1] << endl;
} int count(int a, int b) {int result = 0;for(int i = 0; i < maxn; i++) {if(((a>>i) & 1) != ((b>>i) & 1)) {result++;}}return result;
}

转载于:https://www.cnblogs.com/NJUWZ/p/7093126.html

USACO hamming相关推荐

usaco Hamming Codes
就是在二进制上的枚举,B位的意思就是十进制不超过1<<B,数字很小枚举即可每枚举一个跟前面的进行比较看符不符合题意,就是进行按位异或然后统计其中1的个数大于等于D就符合题意. /* ID: ...
USACO 2.1 海明码 Hamming Codes （模拟+位运算+黑科技__builtin_popcount（n））
题目描述给出 N,B 和 D,要求找出 N 个由0或1组成的编码(1 <= N <= 64),每个编码有 B 位(1 <= B <= 8),使得两两编码之间至少有 D 个单位 ...
USACO Section2.1 Hamming Codes 解题报告【icedream61】
hamming解题报告 ---------------------------------------------------------------------------------------- ...
USACO 做题小结
还记得之前,发过一篇阶段性总结与未来规划..结果由于最近rp爆发(保研成功+进wf)后者显然靠bin神,前者也是运气.因此,放松了一段时间.然后就开始刷usaco了,原因是不用花时间找解题报告在NOC ...
洛谷P1461 海明码 Hamming Codes
P1461 海明码 Hamming Codes 98通过 120提交题目提供者该用户不存在标签USACO 难度普及/提高- 提交讨论题解最新讨论暂时没有讨论题目描述给出 N,B 和 ...
usaco Shaping Regions
这就是usaco 前面的windows area的变形. /* ID:jinbo wu TASK:rect1 LANG:C++ */ #include<iostream> #include ...
usaco Postal Vans(dp)
是哈密顿回路,然后...就不知道怎么写了 ,以前写过类似的不过情况没这么多也没这么复 usaco training 6.1.1 Postal Vans 题解标签: usaco training题解d ...
usaco Beef McNuggets
这两天贼烦,ccf炸了,还有一个烦心事.哎我都不知道自己能不能坚持下去了.马上期末考了.这段时间还是抓紧时间复习吧同时刷usaco的节奏要跟以前一样了,毕竟课少了. 题解: 只要你知道以下的数论结论, ...
usaco前两章小结
usaco 暑假老师有推荐做但是那个题目太长了,而且·大部分都是废话做起来特别慢,而且当时自己基本上什么都不懂,太难了所以看了题解做了两题就放弃了. 转眼就上学了,因为想学习acm所以就胡乱找题做但是 ...