字符串

1.小红的数字串

https://www.nowcoder.com/questionTerminal/d70a5f22b118407f85a88ed7f87ed45c?page=5&onlyReference=false

#include <iostream>
using namespace std;

int main() {
    string s;
    long long k;
    cin >> s;
    cin >> k;
    int n = 0;
    for (long long i = 0; i < s.size(); i++){
        long long m = 0;
        for (long long j = i; j < s.size(); j++) {
           m = m * 10 + s[j] - '0';
           if (m >= k) break; 
           n++;
        }
    }
    cout << n <<endl;
    return 0;
}
// 64 位输出请用 printf("%lld")

 

2.游游的字母串

https://ac.nowcoder.com/acm/contest/66943/B

知识点：

1.    #include <climits>  取极大数     int res = INT_MAX;   LLONG_MAX

2.  取绝对值abs

#include <iostream>
#include <cstdlib>  // for abs(int)
#include <algorithm>
#include <climits>
using namespace std;
int main(){
    string str;
    cin >> str;
    int res = INT_MAX;
    for (int i = 'a'; i <= 'z'; i++){
        int ops = 0;
        for (int j = 0; j < str.size(); j++){
            ops = ops + min(abs(str[j] - i), 26-abs(str[j] - i));    
        }
        res = min(res, ops);
    }
    
    cout << res << endl;
}

 

3.小红的纸牌游戏

https://www.nowcoder.com/practice/a454e716209c467c924577295f6e0d8d?tpId=182&rp=1&sourceUrl=%2Fexam%2Foj&difficulty=2&judgeStatus=&tags=&title=&gioEnter=menu

知识点：

1.find函数 找到返回位置，找不到返回-1

2.%2  偶数是小红操作，奇数是小紫操作

#include<bits/stdc++.h>
using namespace std;
int main() {
    int n, k;
    cin >> n >> k;
    string str;
    cin >> str;
    for (int i = 0; i < n - k; i++) {
        if (i % 2 == 0) {
            int pos = str.find('0');
            if (pos != -1 && pos != str.size() - 1) {
                str = str.substr(0, pos) + str.substr(pos + 1, str.size());
            } else {
                str.pop_back();
            }
        } else {
            int pos = str.find('1');
            if (pos != -1 && pos != str.size() - 1) {
                str = str.substr(0, pos) + str.substr(pos + 1, str.size());
            } else {
                str.pop_back();
            }
        }
    }
    cout << str << endl;
}

 

4.小红的循环移位

https://ac.nowcoder.com/acm/contest/80742/C

知识点：

stoi函数：字符串转整形

 

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

int main() {
    string s;
    cin >> s;
    int n = s.size();

    int ans = -1;

    for (int i = 0; i < n; ++i) {
        // 每次左移一位：把 s[0] 放到末尾
        if (n >= 2) {
            string lastTwo = s.substr(n - 2, 2);
            int num = stoi(lastTwo);
            if (num % 4 == 0) {
                ans = i;
                break;
            }
        } else {
            // 特判长度为1的情况
            int num = s[0] - '0';
            if (num % 4 == 0) {
                ans = i;
                break;
            }
        }

        // 左移一次
        s = s.substr(1) + s[0];
    }

    cout << ans << endl;
    return 0;
}

 

 5.对称之美

https://ac.nowcoder.com/acm/contest/10746/H

#include <bits/stdc++.h>
using namespace std;

bool judge(vector<string>& strs){
    int n = strs.size();
    for(int i=0; i <= n/2; i++){
        if(i == n-i-1){
            break;
        }
        for(int j = 0; j < strs[i].size(); j++){
            if(strs[n-i-1].find(strs[i][j]) != -1){
                break;
            } else {
                if(j == strs[i].size() - 1){
                    return false;
                }
            }
        }
    }
    return true;
}

int main(){
    int t;
    cin >> t;
    while (t--){
        int n;
        cin >> n;
        vector<string> strs;
        while (n--){
            string str;
            cin >> str;
            strs.push_back(str);
        }
        if (judge(strs)) {
            cout << "Yes" << endl;
        } else {
            cout << "No" << endl;
        }
    }
}

 

 

6.小红的01串

https://www.nowcoder.com/questionTerminal/d371489291b8455ab5e7fb86a26ee5bc?f=discussion

要注意如果是01001   不能变成01101 只能变成00001

#include <iostream>
using namespace std;

int main() {
    string str;
    cin >> str;
    int n = 0;
    for (int i = 0; i < str.size() - 3; i++){
        string substr = str.substr(i,3);
        if (substr == "010") {
            if (str.substr(i+2, 3) == "001") {
                str[i+1] = '0';
            } else {
                str[i+2] = '1';
            }
            n++;
        }
        if (substr == "101") {
            if (str.substr(i+2, 3) == "110"){
                str[i+1] = '1';
            } else {
                str[i+2] = '0';
            }
            n++;
        }
    }
    cout << n << endl;
}
// 64 位输出请用 printf("%lld")