Likou 1143
Given two strings text1 and text2, return the length of the longest common subsequence of these two strings.
A subsequence of a character string refers to a new character string: it is a new character string formed by deleting some characters (or no characters) from the original string without changing the relative order of the characters.
For example, "ace" is a subsequence of "abcde", but "aec" is not a subsequence of "abcde". The "common subsequence" of two strings is a subsequence shared by the two strings.
If the two strings have no common subsequence, 0 is returned.
Example 1:
Input: text1 = "abcde", text2 = "ace"
Output: 3
Explanation: The longest common subsequence is "ace", and its length is 3.
Example 2:
Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc", and its length is 3.
Example 3:
Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: Two strings have no common subsequence, return 0.
Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/longest-common-subsequence
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
Idea:
dp[i][j] represents the number of the longest common subsequence in the first i characters of the first character and the first j characters of the second character.
When processing input, you can read each char directly into str.
Code:
class Solution {
public:
int longestCommonSubsequence(string text1, string text2) {
int m = text1.size();
int n = text2.size();
vector<vector<int>> dp(m+1,vector(n+1,0));
for(int i = 1;i <=m;++i){
for(int j = 1;j<=n;++j){
if(text1[i-1] == text2[j-1]){
dp[i][j] = dp[i-1][j-1] + 1;
}
else{
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
}
}
}
return dp[m][n];
}
};
Palindromic prime
There is a positive integer, and hope to get a palindrome prime after removing a certain number in this number. A palindrome prime number means that a prime number is also a palindrome number. Note: One digit is also considered to be a palindrome. Enter two positive integers N and M to satisfy N<M. Please write a program to count how many numbers between N and M satisfy the above requirements?
Idea:
Brute force method, traverse every number in the range, and remove every number in them. Then determine whether it is a palindrome prime. The cumulative number.
Code:
#include <vector>
#include <iostream>
#include <string>
#include <stdlib.h>
#include <cmath>
using namespace std;
class Solution{
public:
int solution(int low, int high){
int ans = 0;
for(int i=low; i<=high; ++i){
if(helper(i)){
++ ans;
}
}
return ans;
}
bool helper(int num){
string str = to_string(num);//整数转字符串
for (int i = 0; i < str.length(); ++i) {
string temp = str;
temp.erase(i,1);
if (isHuiwen(atoi(temp.c_str())) && isSushu(atoi(temp.c_str()))) {
//字符串转整数
return true;
}
}
return false;
}
bool isSushu(int num){
if(num<=1) return false;
int sqt = sqrt(num);
for(int i=2; i<=sqt; ++i){
if(num % i == 0)
return false;
}
return true;
}
//dp 判断回文串
bool isHuiwen(int num){
string s = to_string(num);
int n = s.size();
vector<vector<bool>> Mat(n,vector<bool>(n,false));
for(int i = n-1;i>=0;--i){
for(int j = i;j<n;++j){
if(j ==i){
//从尾部开始，一个字符的时候，一定是回文串
Mat[i][j] =true;
}
else if(j == i+1){
//从尾部判断，两个字符的时候，取决于两个字符是否相等？相等的话就是回文串，不相等的话就不是。
Mat[i][j] = s[i] == s[j];
}
else
Mat[i][j] = (s[i] == s[j]) && (Mat[i+1][j-1]);//i和j位置上的字符相等 && [i+1,j-1]是回文串的话，[i,j]就是回文串
}
}
return Mat[0][n-1];
}
};
int main(){
Solution sol;
int low, high;
while (cin>>low>>high) {
cout << sol.solution(low, high) << endl;
}
}
How to enter a fixed number of decimal places in C++
#include <iomanip>
#include <iostream>
using namespace std;
int main(){
double a = 12.334;
cout<<fixed<<setprecision(1)<<a<<endl;
}