hdu2072 单词数 和 hdu4018 Parsing URL

本文通过两个具体的编程案例展示了如何使用C++解析URL中的域名部分。案例包括了读取输入字符串、定位URL中的特定字符并提取域名的过程。适用于初学者理解基本的字符串操作及编程逻辑。

hdu 2072 单词数

#include <iostream>
#include <iostream>
using namespace std;

#define N 10000
char str[N];
char url[N];


int main()
{
	int i, n, t, k, m = 1;

	scanf("%d", &t);
	getchar();
	while(t--)
	{	
		gets(str);
		n = strlen(str);
		k = 0;
		i = 0;
		while(i < n && str[i] != '/') ++i;
		i += 2;
		for (;i < n && str[i] != '/' && str[i] != ':'; ++i)
		{
			url[k++] = str[i];
		}
		url[k] = 0;
		printf("Case #%d: %s\n", m++, url);
	}
	return 0;
}

hdu4018 Parsing URL

#include <iostream>
using namespace std;

#define N 10000
char str[N];
char url[N];


int main()
{
	int i, n, t, k, m = 1;

	scanf("%d", &t);
	getchar();
	while(t--)
	{	
		gets(str);
		n = strlen(str);
		k = 0;
		i = 0;
		while(i < n && str[i] != '/') ++i;
		i += 2;
		for (;i < n && str[i] != '/' && str[i] != ':'; ++i)
		{
			url[k++] = str[i];
		}
		url[k] = 0;
		printf("Case #%d: %s\n", m++, url);
	}
	return 0;
}



### HDU OJ 1063 Problem Solution in C Language The provided code snippet appears to be incomplete and does not directly correspond to the specific requirements of HDU OJ 1063. For solving problems on platforms like HDU OJ, it is crucial to understand both the problem statement thoroughly and apply appropriate algorithms or data structures. For HDU OJ 1063 titled "Fill Blank," which involves filling blanks based on given conditions, an efficient approach can involve dynamic programming techniques combined with careful input parsing and output formatting[^1]. Below is a more structured way to tackle this kind of problem using C: ```c #include <stdio.h> #include <string.h> #define MAXN 1005 // Assuming maximum length as per constraints char str[MAXN]; int dp[MAXN][MAXN]; void solve(int n) { memset(dp, 0, sizeof(dp)); for (int len = 2; len <= n; ++len) { // Length from smallest non-trivial case upwards for (int i = 0; i + len - 1 < n; ++i) { int j = i + len - 1; if (str[i] == '(' && str[j] == ')') { dp[i][j] = dp[i + 1][j - 1] + 2; } for (int k = i; k < j; ++k) { dp[i][j] = ((dp[i][j]) > (dp[i][k] + dp[k + 1][j])) ? (dp[i][j]) : (dp[i][k] + dp[k + 1][j]); } } } printf("%d\n", dp[0][n - 1]); // Output result according to sample outputs } int main() { while (~scanf("%s", str)) { int n = strlen(str); solve(n); // Process each test case individually } return 0; } ``` This program reads strings composed mainly of parentheses `(` and `)` characters, calculates how many pairs are correctly matched by applying dynamic programming principles, and prints out results accordingly. The core idea lies within maintaining a two-dimensional array where `dp[i][j]` represents the longest valid substring between positions `i` and `j`.
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