string算法刷题

2022-08-26

[toc]

类型转换

char
1
2
string a;
char n[10] = a.c_str();

int

1 2	string a; int n = atoi(a.c_str());

1
2
3

string s;
cin >> s;
float n = stof(s);

1 2	string str; int n = stoi(str);

double
1
2
string a;
double n = atof(a.c_str());
to string
1
2
int a;
string s = to_string(a);

字符串输入

getline(cin, str);

前面要是有cin和scanf,一定要用getchar()来把末尾的回车读取.

include 函数

isalpha() 是字母返回true
islower() 是小写字母返回true
isupper() 是大写字母返回true
isdigit() 是数字返回true
tolower() 转小写字母
toupper() 转大写字母

1024 Palindromic Number

思路：开始时候是用字符串存储，但是相加的操作又转为了int，发现会范围溢出，因此add操作也改为string相加，即为大整数相加。

#include <stdio.h>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;

string s;
void add(string str){
    int len = s.length();
    int carry = 0;
    for (int i = len - 1; i >= 0; i--) {
        s[i] = s[i] + str[i] + carry - '0';
        carry = 0;
        if (s[i] > '9') {
            s[i] = s[i] - 10;
            carry = 1;
        }
    }
    if (carry) {
        s = "1" + s;
    }
}

int main (int argc, char *argv[])
{
    int k;
    cin >> s >> k;
    int i;
    for (i = 0; i <= k; i++) {
        string t = s;
        reverse(t.begin(), t.end());
        if (t == s || i == k) {
            break;
        }
        add(t);
    }
    cout << s << endl << i;
    return 0;
}

注意：add函数s[i]计算的时候加上一个int值，最终还是个string类型，因此比较的时候要和‘9’进行比较，以及进位。

1001 A+B Format

思路：将a+b的和用string进行处理，截取字符串操作。

#include <cstdio>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;

int main (int argc, char *argv[])
{
    int a, b;
    scanf("%d%d", &a, &b);
    int sum = a + b;
    string str = to_string(sum);
    bool flag = false;
    if (str[0] == '-') {
        flag = true;
        str.erase(str.begin());
    }
    if (str.length() < 4) {
        if (flag) {
            printf("-");
        }
        cout << str << endl;
    }
    else if (str.length() < 7){
        int i = str.length() - 3;
        if (flag) {
            printf("-");
        }
        cout << str.substr(0, i) << "," << str.substr(i, 3) << endl;
    }
    else {
        if (flag) {
            printf("-");
        }
        cout << str.substr(0, 1) << "," << str.substr(1, 3) << "," << str.substr(4,3)<< endl;
    }
    return 0;
}

注意：可能有两个逗号的情况，位数大于六位。

1077 Kuchiguse

思路:用reverse()翻转string,然后每一个string进行判断最长前缀.

#include <cstdio>
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <string.h>
using namespace std;

int main (int argc, char *argv[])
{
    int n;
    scanf("%d\n", &n);
    string suffix, str;
    getline(cin, suffix); 
    reverse(suffix.begin(), suffix.end());
    for (int i = 0; i < n - 1; i++) {
        getline(cin, str); 
        reverse(str.begin(), str.end());
        for (int j = 0; j < suffix.length(); j++) {
            if (suffix[j] != str[j]) {
                suffix = suffix.substr(0, j);
                break;
            }            
        }
    }
    if (suffix.size() == 0) {
        cout << "nai" << endl;
    }
    else {
        for (int i = suffix.size() - 1; i >= 0; i--) {
            cout << suffix[i];
        }
        cout << endl;
    }
    return 0;
}

1035 password

思路：用一个map进行匹配即可。

#include <cstdio>
#include <stdio.h>
#include <algorithm>
#include <map>
#include <string>
#include <vector>
#include <iostream>
using namespace std;
const int N = 1010;

vector<string> vpwd;
vector<string> vname;
map<char, char> m;
int Index = 0;
void change(string name, string &str){
    bool flag = false;
    int len = str.length();
    for (int i = 0; i < len; i++) {
        if (str[i] == '1' || str[i] == '0' || str[i] == 'l' || str[i] == 'O') {
            str[i] = m[str[i]];
            flag = true;
        }
    }
    if (flag) {
        Index++;
        vname.push_back(name);
        vpwd.push_back(str);
    }
}

int main (int argc, char *argv[])
{
    m['1'] = '@';
    m['0'] = '%';
    m['l'] = 'L';
    m['O'] = 'o';
    int n;
    scanf("%d", &n);
    string name, password;
    for (int i = 0; i < n; i++) {
        cin >> name >> password;
        change(name, password);
    }
    if (Index == 0 && (n == 1 || n == 0)) {
        printf("There is %d account and no account is modified\n", n);
    }
    else if (Index == 0 && n != 1 && n != 0) {
        printf("There are %d accounts and no account is modified\n", n);
    }
    else if (Index != 0) {
        printf("%d\n", Index);
        for (int i = 0; i < Index; i++) {
            cout << vname[i] << " " << vpwd[i] << endl;
        }
    }
    return 0;
}

注意输出的格式is和are以及accounts单复数。

1061 Dating

思路：什么垃圾题目，又臭又长细节贼多。

#include <cstdio>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <iostream>
#include <map>
using namespace std;

map<char, string> m;
int gethour(char a){
    if (a >= 'A' && a <= 'N') {
        return a - 'A' + 10;
    }
    else {
        return a - '0';
    }
}
int main (int argc, char *argv[])
{
    m['A'] = "MON";
    m['B'] = "TUE";
    m['C'] = "WED";
    m['D'] = "THU";
    m['E'] = "FRI";
    m['F'] = "SAT";
    m['G'] = "SUN";
    char d, h;
    int mi;
    string str1, str2, str3, str4;
    cin >> str1 >> str2 >> str3 >> str4;
    int i = 0, j = 0;
    while (i < str1.length() && i < str2.length()) {
        if (str1[i] == str2[i] && str1[i] >= 'A' && str1[i] <= 'G') {
            d = str1[i];
            break;
        } 
        i ++;
    }
    i++ ;
    while (i < str1.length() && i < str2.length()) {
        if (str1[i] == str2[i] && ((str1[i] >= 'A' && str1[i] <= 'N') || (str1[i] >= '0' && str1[i] <= '9'))) {
            h = str1[i];
            break;
        } 
        i ++;
    }
    while (j < str3.length() && j < str4.length()) {
        if (str3[j] == str4[j] && ((str3[j] >='a' && str3[j] <= 'z') || (str3[j] >='A' && str3[j] <= 'Z'))) {
            mi = j;
            break;
        } 
        j ++;
    }
    cout << m[d] << " ";
    printf("%02d:%02d\n", gethour(h), mi);
    return 0;
}

Scientific Notation

思路：判断正负，前后补零即可，细节注意。

#include <stdio.h>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;

bool flag = true;
int main (int argc, char *argv[])
{
    string str;
    string source;
    string last;
    cin >> str;
    if (str[0] == '-') {
        flag = false;
    }
    str.erase(str.begin());
    str.erase(str.begin() + 1);
    int len = str.find('E');
    source = str.substr(0, len);
    last = str.substr(len + 1, str.length() - len - 1);
    if (last[0] == '-') {
        last.erase(last.begin());
        int n = stoi(last);
        for (int i = 0; i < n; i++) {
            source = '0' + source;
        }
        source.insert(1, ".");
    }
    else {
        string s = "";
        last.erase(last.begin());
        int n = stoi(last);
        s = s + source[0];
        int i;
        for (i = 1; i <= n && i < source.length(); i++) {
            s = s + source[i];
        }
        if (i == source.length()) {
            for (int j = 0; j < n - source.length() + 1; j++) {
                s += "0";
            }
        }
        else {
            s += ".";
            for (int j = i; j < source.length(); j++) {
                s += source[j];
            }
        }
        source = s;
    }
    if (!flag) {
        source = "-" + source;
    }
    cout << source << endl;
    return 0;
}

注意：后面补零，可能还是小数，如：+1.234234234E+05 结果为123423.4234，并没有补零。