String Title Collection

8. String matching

Give two strings a and b of the same length.

If the character a[i] on string a is the same as the character b[i] on string b at a certain position ii, the character at this position is matched.

If the ratio of the number of matching positions of two strings to the total length of the strings is greater than or equal to k, the two strings are said to match.

Now please judge whether the given two strings match.

The first line contains a floating-point number k, the second line contains string a, and the third line contains string b.

The string you entered does not contain spaces.

If the two strings match, yes is output.

Otherwise, no is output.

0 ≤ k ≤ 1 ,
The length of the string does not exceed 100.

0.4 abcde xbacd

no

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { string a,b; double k; cin>>k>>a>>b; int cnt=0; for(int i=0;i<a.size();i++) { if(a[i]==b[i]) { cnt++; } } if((double)cnt/a.size()>=k) { printf("yes"); } else printf("no"); return 0; }

9. Ignore case and compare string size

Generally, we can use strcmp to compare the size of two strings. The comparison method is to compare the two strings from front to back character by character (according to the size of ASCII code value) until different characters appear or \ 0 is encountered.

If all characters are the same, they are considered the same; If different characters appear, the comparison result of the first different character shall prevail.

But sometimes, when we compare the size of strings, we want to ignore the size of letters. For example, hello and hello are equal when ignoring the case of letters.

Please write a program to compare the case of two strings ignoring the case of letters.

The input is two lines, one string per line, a total of two strings. Note that the string may contain spaces.

The data guarantees that the length of each string does not exceed 80.

If the first string is smaller than the second string, output a character <.

If the first string is larger than the second string, output a character >.

If two strings are equal, output a character =.

Hello hello

=

#include<iostream> #include<cstring> #include<cstdio> using namespace std; int main() { char a[100],b[100]; fgets(a,100,stdin); fgets(b,100,stdin); if(a[strlen(a)-1]=='\n')a[strlen(a)-1]=0;//Remove the carriage return at the end if(b[strlen(b)-1]=='\n')b[strlen(b)-1]=0;//Remove the carriage return at the end for(int i=0;a[i];i++) { if(a[i]>='A'&&a[i]<='Z') { a[i]+=32; } } for(int i=0;b[i];i++) { if(b[i]>='A'&&b[i]<='Z') { b[i]+=32; } } int t=strcmp(a,b); if(t==0) puts("="); else if(t<0) puts("<"); else puts(">"); return 0; }

10. Remove redundant spaces

Enter a string, which may contain multiple consecutive spaces. Please remove the extra spaces and leave only one space.

A line containing a string.

Output the string after removing redundant spaces, accounting for one line.

The length of the input string does not exceed 200.
Ensure that there are no spaces at the beginning and end of the input string.

Hello world.This is c language.

Hello world.This is c language.

Here is a tip that can make full use of the features of cin and cout, so easy!

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { string a; while(cin>>a) { cout<<a<<' '; } return 0; }

11. Information encryption

In the process of transmitting information, in order to ensure the security of information, we need to encrypt the original information to form encrypted information, so that the information content will not be stolen by the listener.

Now give a string and encrypt it.

The encryption rules are as follows:

1. The lowercase letter in the string, a is encrypted as b, b is encrypted as c,..., y is encrypted as z, and z is encrypted as a.
2. The uppercase letter in the string. A is encrypted as B, B is encrypted as C,..., Y is encrypted as Z, and Z is encrypted as a.
3. Other characters in the string are not processed.

Please output the encrypted string.

A line containing a string. Note that the string may contain spaces.

Output encrypted string.

The length of the input string does not exceed 100.

Hello! How are you!

Ifmmp! Ipx bsf zpv!

#include<iostream> #include<cstdio> #include<string> #include<algorithm> using namespace std; int main(){ string s1; getline(cin,s1); //cout << s1.size()<<endl; for(int i =0;i<s1.size();i++){ if(s1[i] == 'Z') s1[i]='A'; else if (s1[i] == 'z') s1[i]='a'; else if((s1[i] >= 'a' && s1[i] <= 'z')||(s1[i] >= 'A' &&s1[i] <= 'Z')) s1[i] = (char)((int)s1[i]+1); } cout << s1 <<endl; return 0; }

12. Output string

Given a string a, please output string b according to the following requirements.

Given the ASCII value of the first character of string a plus the ASCII value of the second character, the first character of b is obtained;

Given the ASCII value of the second character of string a plus the ASCII value of the third character, the second character of b is obtained;

...

Given the ASCII value of the penultimate character of string a plus the ASCII value of the last character, the penultimate character of b is obtained;

Given the ASCII value of the last character of string a plus the ASCII value of the first character, the last character of b is obtained.

Enter a line containing the string a. Note that the string may contain spaces.

The data guarantees that the ASCII value of characters in the string does not exceed 63.

The output is a line containing the string b.

Length of 2 ≤ a ≤ 100

1 2 3

QRRSd

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { string a; getline(cin,a); for(int i=0;i<a.size()-1;i++) { cout<<char(a[i]+a[i+1]); } cout<<char(a[a.size()-1]+a[0]); return 0; }

13. Word replacement

Enter a string that ends with a carriage return (the string length does not exceed 100).

The string consists of several words separated by a space. All words are case sensitive.

Now you need to replace one word with another and output the replaced string.

Enter a total of 3 lines.

The first line is a string containing multiple words s;

The second line is the word a to be replaced (no longer than 100);

Line 3 is the word B (no longer than 100) where a will be replaced.

A total of one line, output the string after replacing all words a in s with b.

You want someone to help you You I

I want someone to help you

Replace a string with a character stream

Just use ssin as a cin

#include <sstream> stringstream ssin(s);

#include<iostream> #include<sstream> using namespace std; int main() { string s,a,b; getline(cin,s); cin>>a>>b; stringstream ssin(s); string str; while(ssin>>str) { if(str==a) cout<<b<<' '; else cout<<str<<' '; } return 0; }

14. The longest consecutive character in a string

Find the longest consecutive character in a string and output the character and its occurrence times. There are no white space characters (space, carriage return and tab) in the string. If there is more than one such character, the first one will be output.

In the first line, enter the integer N, which represents the number of groups of test data.

Each group of data occupies one line and contains a string without white space characters. The length of the string shall not exceed 200.

A total of one line, output the longest consecutive characters and their occurrence times, separated by spaces.

2 aaaaabbbbbcccccccdddddddddd abcdefghigk

d 10 a 1

#include<iostream> using namespace std; int main() { int n; cin>>n; while(n--) { string str; cin>>str; int cnt=0; char c; for(int i=0;i<str.size();i++)//The first kind of double pointer algorithm { int j=i; while(j<str.size()&&str[j]==str[i]) j++; if(j-i>cnt) { cnt=j-i; c=str[i]; } i=j-1; } cout<<c<<' '<<cnt<<endl; } return 0; }

15. Longest word

A simple English sentence ending with. Separated by spaces between words without abbreviations and other special forms. Find the longest word in the sentence.

Enter this simple English sentence with a length of no more than 500.

The longest word in the sentence. If more than one, the first one is output.

I am a student of Peking University.

University

#include<iostream> using namespace std; int main() { string res,str; while(cin>>str) { if(str.back()=='.') str.pop_back(); if(str.size()>res.size()) res=str; } cout<<res<<endl; return 0; }

16. Inverted words

Write a program, read in a line of English (only letters and spaces, separated by a single space between words), reverse the order of all words and output, still separated by a single space.

Input as a string (string length up to 100).

The output is a string sorted as required.

I am a student

student a am I

#include<iostream> using namespace std; int main() { string str[100]; int n=0; while(cin>>str[n]) n++; for(int i=n-1;i>=0;i--) { cout<<str[i]<<' '; } cout<<endl; return 0; }

17. String shift inclusion problem

For a string, define a cyclic shift operation as: move the first character of the string to the end to form a new string.

Given two strings s1 and s2, it is required to determine whether one string is a substring of a new string after the other string is shifted several times.

For example, CDAA is a substring of BCDAA, a new string generated after two shifts of AABCD, while ABCD and ACBD cannot obtain a substring of the new string through multiple shifts.

A line containing two strings separated by a single space.

The string contains only letters and numbers and is no longer than 30.

If a string is a substring of a new string generated by several cyclic shifts of another string, true is output; otherwise, false is output.

AABCD CDAA

true

#include <iostream> #include <algorithm> using namespace std; int main() { string a, b; cin >> a >> b; if (a.size() < b.size()) swap(a, b); for (int i = 0; i < a.size(); i ++ ) { a = a.substr(1) + a[0]; for (int j = 0; j + b.size() <= a.size(); j ++ ) { int k = 0; for (; k < b.size(); k ++ ) if (a[j + k] != b[k]) break; if (k == b.size()) { puts("true"); return 0; } } } puts("false"); return 0; }

18. String power

Given two strings a and b, we define a × b is their connection.

For example, if a=abc and b=def, then a × b=abcdef.

If we consider the connection as multiplication, the power of a nonnegative integer will be defined in the usual way: a0 = ` ` (empty string), a(n+1)=a × (an).

The input contains multiple groups of test samples, each of which occupies one line.

Each set of samples contains a string s, whose length does not exceed 100.

The final test sample will be followed by a dot as a line.

For each s, you need to output the maximum n so that there is a string a, let s=an.

abcd aaaa ababab .

1 4 3

#include<iostream> using namespace std; int main() { string str; while(cin>>str,str!=".") { int len=str.size(); for(int n=len;n;n--) { if(len%n==0) { int m=len/n; string s=str.substr(0,m); string r; for(int i=0;i<n;i++) { r+=s; } if(r==str) { cout<<n<<endl; break; } } } } return 0; }

19. Maximum span of string

There are three strings S, S1 and S2, where the length of s does not exceed 300 and the length of S1 and S2 does not exceed 10.

Now, we want to detect whether S1 and S2 appear in s at the same time, and S1 is located on the left of S2 and does not cross each other in S (that is, the right boundary point of S1 is on the left of the left boundary point of S2).

Calculate the maximum span that meets the above conditions (i.e., the maximum spacing: the number of characters between the starting point of the rightmost S2 and the ending point of the leftmost S1).

If there is no S1 and S2 that meet the conditions, output − 1.

For example, S= abcd123ab888efghij45ef67kl, S1= ab, S2= ef, where S1 occurs twice in s, S2 also occurs twice in s, and the maximum span is 18.

Enter a line containing three strings S, S1 and S2 separated by commas.

The data guarantees that the three strings do not contain spaces and commas.

Outputs an integer representing the maximum span.

If no S1 and S2 that meet the conditions exist, output − 1.

abcd123ab888efghij45ef67kl,ab,ef

18

#include<iostream> using namespace std; int main() { string s,s1,s2; char c; while(cin>>c,c!=',') { s+=c; } while(cin>>c,c!=',') { s1+=c; } while(cin>>c) { s2+=c; } if(s.size()< s1.size()||s.size()<s2.size()) { puts("-1"); } else { int l=0; while(l+s1.size()<=s.size()) { int k=0; while(k<s1.size()) { if(s[l+k]!=s1[k]) { break; } else k++; } if(k==s1.size()) break; else l++; } int r = s.size() - s2.size(); while (r >= 0) { int k = 0; while (k < s2.size()) { if (s[r + k] != s2[k]) break; k ++ ; } if (k == s2.size()) break; r -- ; } l += s1.size() - 1; if (l >= r) puts("-1"); else printf("%d\n", r - l - 1); } return 0; }

20. Longest common string suffix

Give several strings and output the longest common suffix of these strings.

It consists of several sets of inputs.

The first line of each set of inputs is an integer N.

When N is 0, the input ends, otherwise there will be N lines of input, and each line is a string (the string does not contain blank characters).

The length of each string shall not exceed 200.

A total of one line, the longest common suffix of N strings (may be empty).

1≤N≤200

3 baba aba cba 2 aa cc 2 aa a 0

ba a

#include <iostream> using namespace std; const int N = 200; int n; string str[N]; int main() { while (cin >> n, n) { int len = 1000; for (int i = 0; i < n; i ++ ) { cin >> str[i]; if (len > str[i].size()) len = str[i].size(); } while (len) { bool success = true; for (int i = 1; i < n; i ++ ) { bool is_same = true; for (int j = 1; j <= len; j ++ ) if (str[0][str[0].size() - j] != str[i][str[i].size() - j]) { is_same = false; break; } if (!is_same) { success = false; break; } } if (success) { break; } else len -- ; } cout << str[0].substr(str[0].size() - len) << endl; } return 0; }

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