数字IC笔试千题解--编程&&脚本篇（八）

 
#include 
 
void main(){
    char str[100] = {0};
    int count[26] = {0};
 
    int i = 0;
    int count_index;
 
    scanf("%s", str);
 
    while(str[i] != '\0'){
        count_index = (str[i]- 'a');
        count[count_index] = count[count_index] + 1;
        i++;
    }
    for(i=0;i<26;i++){
        if(count[i]!=0){
            printf("%d%c",count[i],'a'+i);
        }
    }
    printf("\n");
}

 
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
 
 
class Solution:
 
    def maxDepth(self, root):
        length = 0
 
        def dfs(root, d):
            nonlocal length
            if not root:
                return
            length = max(length,d)
            dfs(root.left, d+1)
            dfs(root.right, d+1)
 
        dfs(root, 1)
        return length
 
 
if __name__ == '__main__':
    # 定义题中所给的树
    root = TreeNode(3)
    root.left = TreeNode(9)
    node = TreeNode(20)
    root.right = node
    node.left = TreeNode(15)
    node.right = TreeNode(7)
 
    s = Solution()  # 例化类，才能调用类中的类函数
    print(s.maxDepth(root))

 
def calculate(s: str) -> int:
    num = 0     # 记录当前数字
    sign = 1    # 记录当前符号
    stack = []  # 用来存储当前括号外的结果和符号
    res = 0     # 用来记录最终结果
 
    for i in range(len(s)):
        if s[i].isdigit():
            num = num * 10 + int(s[i])
        elif s[i] == '+':
            res += num * sign
            num = 0
            sign = 1
        elif s[i] == '-':
            res += num * sign
            num = 0
            sign = -1
        elif s[i] == '(':
            stack.append(res)    # 先存结果，再存数字
            stack.append(sign)    
            res = 0
            sign = 1
        elif s[i] == ')':
            res += num * sign
            num = 0
            res *= stack.pop()
            res += stack.pop()
        
    return res + num * sign
 
if __name__ == '__main__':
    s = "(1+(4+5+2)-3)+(6+8)"
    print(calculate(s))

 
import os
 
def find_fail_lines(dir_path):
    try:
        with open(os.path.join(dir_path, 'result.log'), 'r') as f1, open(os.path.join(dir_path, 'fail.log'), 'w') as f2:
            fail_lines = [line for line in f1 if 'fail' in line.lower()]
            f2.writelines(fail_lines)
            print('Successfully saved fail lines to fail.log')
    except FileNotFoundError:
        print('result.log file not found in the directory')
 
# example usage
find_fail_lines('E:/modelsim_project/file_handle/')

 
with open('file.txt') as file:
    for i, line in enumerate(file, 1):
        print(f"Line {i}: {line}",end="")

 
keywords = {}  # 创建一个空字典，用于保存匹配到的关键字和行号
with open('file.txt') as file:
    for i, line in enumerate(file, 1):
        for j in range(1, 100):
            keyword = f'biren{j:02d}'  # 生成关键字，如 'biren01'、'biren02' 等
            if keyword in line:
                if i not in keywords:  # 如果当前行号还没有在字典中出现过，则添加新的键值对
                    keywords[i] = [keyword]
                else:  # 如果当前行号已经在字典中存在，则将匹配到的关键字添加到已有的列表中
                    keywords[i].append(keyword)
for key, value in keywords.items():
    print(f"行号 {key},匹配到的字符为：{value}")

 
#include 
 
int is_prime(int num) {
    if (num <= 1) {
        return 0;
    }
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) {
            return 0;
        }
    }
    return 1;
}
 
int main() {
    int count = 0;
    printf("素数有：");
    for (int i = 100; i <= 1000; i++) {
        if (is_prime(i)) {
            count++;
        }
    }
    printf("共有%d个素数\n", count);
    return 0;
}

 
a = [1,1,2,4,5]
b = [1,1,2,3,7]
intersection_A_B = set(a)-(set(a)-set(b))
union_A_B = set(a).union(set(b))
print(intersection_A_B)
print(union_A_B)

 
#include 
int main(void){
    int matrix[3][4] = {{0,1,2,3},{8,7,6,5},{-5,9,-3,4}};
    int row,column,max;
    max = 0;
    for (int i =0;i<=2;i++){
        for(int j=0;j<=3;j++){
            if(matrix[i][j]>=max){
                row = i;
                column = j;
                max = matrix[i][j];
            }
        }
    }
    printf("max value is:%d,the row is:%d,the column is:%d\n",matrix[row][column],row+1,column+1);
}