AOJ基础题 ITP1_9_D Transformation

Transformation

Write a program which performs a sequence of commands to a given string str. The command is one of:

print a b: print from the a-th character to the b-th character of str
reverse a b: reverse from the a-th character to the b-th character of str
replace a b p: replace from the a-th character to the b-th character of str with p

Note that the indices of str start with 0.

Input

In the first line, a string str is given. str consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.

Output

For each print command, print a string in a line.

Constraints

1≤ length of str ≤ 1000
1 ≤ q ≤ 100
0 ≤ a ≤ b < length of str
for replace command, b −a + 1= length of p

Sample Input

xyz
3
print 0 2
replace 0 2 abc
print 0 2

Sample Output

xyz
abc

問題を解く

最简单解法，reverse这里依然选择换下标。

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

char str[1001];
char re[1001];

char c1[] = "replace";
char c2[] = "reverse";
char c3[] = "print";

void replace(int a, int b)
{
    int count = 0;
    for (int i = a; i < b+1; i++)
    {
        str[i] = re[count];
        count++;
    }
}

void reverse(int a, int b)
{
    char temp = 0;
    int n = 0;
    if ((b - a + 1) % 2 == 0)
    {
        for (int i = a; i < ((b+a+1)/2); i++)
        {
            temp = str[i];
            str[i] = str[b - n];
            str[b - n] = temp;
            n++;
        }
    }
    else if ((b - a + 1) % 2 == 1)
    {
        for (int i = a; i < ((b+a+2) / 2); i++)
        {
            temp = str[i];
            str[i] = str[b - n];
            str[b - n] = temp;
            n++;
        }
    }

}

void print(int a, int b)
{
    for (int i = a; i < b + 1; i++)
    {
        cout << str[i];
    }
    cout << endl;
}


int main()
{
    int n = 0;
    int a = 0;
    int b = 0;

    cin >> str;
    cin >> n;

    for (int i = 0; i < n; i++)
    {
        char c[10];
        cin >> c;


        if (strcmp(c, c1) == 0)
        {
            cin >> a >> b >> re;
            replace(a, b);
        }
        else if (strcmp(c, c2) == 0)
        {
            cin >> a >> b;
            reverse(a, b);
        }
        else if (strcmp(c, c3) == 0)
        {
            cin >> a >> b;
            print(a, b);
        }
    }

    return 0;
}

A Good Try

如果使用string类来写就太简单了，substr()、replace()、reverse()这几个方法信手拈来。

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

int main() {
    string str;
    int q;
    cin >> str;
    cin >> q;

    while (q--) {
        string s;
        cin >> s;
        if (s == "print") {
            int a, b;
            cin >> a >> b;
            cout << str.substr(a, b - a + 1) << endl;
        }
        else if (s == "reverse") {
            int a, b;
            cin >> a >> b;
            reverse(str.begin() + a, str.begin() + b + 1);
        }
        else {
            int a, b;
            string p;
            cin >> a >> b >> p;
            str.replace(a, b - a + 1, p);
        }
    }
    return 0;
}