一个简单的英语C语言问题

Question

题目描述
Penney’s game is a simple game typically played by two players. One version of the game calls for each player to choose a unique three-coin sequence such as HEADS TAILS HEADS (HTH) . A fair coin is tossed sequentially some number of times until one of the two sequences appears. The player who chose the first sequence to appear wins the game.
For this problem, you will write a program that implements a variation on the Penney Game. You will read a sequence of 40 coin tosses and determine how many times each three- coin sequence appears. Obviously there are eight such three-coin sequences: TTT , TTH , THT , THH , HTT , HTH , HHT and HHH . Sequences may overlap. For example, if all 40 coin tosses are heads, then the sequence HHH appears 38 times.
输入
The first line of input contains a single integer P , (1<=P<=1000), which is the number of data sets that follow. Each data set consists of 2 lines. The first line contains the data set number N . The second line contains the sequence of 40 coin tosses. Each toss is represented as an upper case H or an upper case T , for heads or tails, respectively. There will be no spaces on any input line.
输出
For each data set there is one line of output. It contains the data set number followed by a single space, followed by the number of occurrences of each three-coin sequence, in the order shown above, with a space between each one. There should be a total of 9 space separated decimal integers on each output line.
样例输入
4
1
HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
2
TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
3
HHTTTHHTTTHTHHTHHTTHTTTHHHTHTTHTTHTTTHTH
4
HTHTHHHTHHHTHTHHHHTTTHTTTTTHHTTTTHTHHHHT

样例输出
1 0 0 0 0 0 0 0 38
2 38 0 0 0 0 0 0 0
3 4 7 6 4 7 4 5 1
4 6 3 4 5 3 6 5 6

麻烦吧程序写给我下。

Answer 1

#include <iostream>

#include <cstring>

using namespace std;

class String

{

char *s;

int size;

public:

String(){

size=0;

s = new char[size+1];

s[0]='\0';

}//default constructor, create an empty string, e.g. *s = ""

~String(){

delete [] s;

} //destructor, destroy object

String(const char str[]){

size=strlen(str);

s = new char[size+1];

s[0]='\0';

strcpy(s,str);

}//parameterized constructor

String (const String & str){

size=str.size;

s = new char[size+1];

s[0]='\0';

strcpy(s,str.s);

}//copy constructor

String(int num, const char c){

size=num;

int i;

s = new char[size+1];

for(i=0;i<size;i++){

s[i]=c;

}

s[size]='\0';

}//copy char c num times

String& concate(const String str);//concatenate two strings, append str to current string

int strLength(); //return size of string

//Copies the portion of str that begins at the character position begin and spans len characters

//or until the end of str, if either str is too short or if len is -1

String (const String& str, int begin, int len= -1){

size=strlen(str.s)-begin;

if(len!=-1&&len<size){

size=len;

}

s = new char[size+1];

int i=0;

while(i<size){

s[i]=str.s[begin+i];

i++;

}

s[i]='\0';

}

//Searches the string for the first occurrence of str

//When pos is specified, the search only includes characters at or after position pos

//ignoring any possible occurrences that include characters before pos.

int find(const char str[], int pos,int l);

//operand overloading, already defined for you

friend ostream& operator << (ostream &cout, String str)

{

cout<<str.s;

return (cout);

}

};

String& String:: concate(const String str){

size+=str.size;

strcat(s,str.s);

return *this;

}

int String:: strLength(){

return strlen(s);

}

int String:: find(const char str[], int pos = 0,int l=3){

if(str[0]==s[pos] && str[1]==s[pos+1] &&str[2]==s[pos+2] ){

return 1;

}

else{

return -1;

}

};

int main(){

int n;

cin>>n;

int i;

const char b[8][4]={"TTT","TTH","THT","THH","HTT","HTH","HHT","HHH"};

for(i=1;i<=n;i++){

int d;

cin>>d;

char *p=new char[41];

cin>>p;

String a(p);

int c[8]={0};

int x;

for(x=0;x<39;x++){

int y;

for(y=0;y<8;y++){

if(a.find(b[y],x,3)!=(-1)){

c[y]++;

}

}

}

cout<<i;

for(x=0;x<8;x++){

cout<<" "<<c[x];

}

cout<<endl;

}

return 0;

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