codeforces 610A Pasha and Stick

探讨Pasha如何将棍子切割成四部分以形成矩形而非正方形的方法数量。文章提供了一个简洁的C++代码示例,展示了解决这一数学问题的具体实现。

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A. Pasha and Stick
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Pasha has a wooden stick of some positive integer length n. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be n.

Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.

Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer x, such that the number of parts of length x in the first way differ from the number of parts of length x in the second way.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 2·109) — the length of Pasha's stick.

Output

The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.

Examples
input
Copy
6
output
Copy
1
input
Copy
20
output
Copy
4
Note

There is only one way to divide the stick in the first sample {1, 1, 2, 2}.

Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.


简单的数学题,需要注意的当n为奇数时,输出为0

#include<bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
const int inf = 0x3f3f3f3f;  
int main()   
{  
	// freopen("shuju.txt","r",stdin);
    int n;
    cin>>n;
    if(n&1)
    	printf("0\n");
    else
    {
	    n=n/2;
	    if(n&1)
	        printf("%d\n",n/2);
	    else
	    	printf("%d\n",n/2-1);     
	}
    return 0;  
}



虽然给定引用中未直接提及“Kuroni and Simple Strings”题目的详细信息,但通常这类题目可能与字符串处理、括号匹配等相关。一般而言,题目可能会给一个由括号组成的字符串,要求找能移除的最大数量的不相交的合法括号对,并输移除这些括号对后的相关信息。 ### 解法分析 #### 栈解法 栈解法是处理括号匹配问题的经典方法。通过遍历字符串,将左括号压入栈中,遇到右括号时,若栈顶为左括号,则将栈顶元素弹,表示这是一对匹配的括号。 ```python s = input() stack = [] pairs = [] for i, char in enumerate(s): if char == '(': stack.append(i) else: if stack: left_index = stack.pop() pairs.append((left_index + 1, i + 1)) if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` #### 双指针解法 双指针解法从字符串的两端向中间遍历,分别使用两个指针 `left` 和 `right`。`left` 指针从左向右寻找 `(`,`right` 指针从右向左寻找 `)`,当找到一对匹配的括号时,将它们标记为已移除,继续寻找下一对匹配的括号,直到无法再找到匹配的括号为止。 ```python s = input() n = len(s) left = 0 right = n - 1 pairs = [] while left < right: while left < right and s[left] != '(': left += 1 while left < right and s[right] != ')': right -= 1 if left < right: pairs.append((left + 1, right + 1)) left += 1 right -= 1 if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` ### 复杂度分析 - **栈解法**:时间复杂度为 $O(n)$,其中 $n$ 是字符串的长度。空间复杂度为 $O(n)$,主要用于栈的空间开销。 - **双指针解法**:时间复杂度为 $O(n)$,空间复杂度为 $O(n)$,主要用于存储匹配的括号对。
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