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九度OJ 1006 ZOJ

#include <iostream>#include <string>using namespace std;int getO(string str,int &start,char ch){    int count=0;    for(;start<str.length();start++)    {        if(str[start] == o)        {           count++;        }        else if(str[start]==ch)        {            start++;            return count;        }        else            return INT_MIN;    }    return INT_MIN;}int main(){    string str;    while(cin>>str)    {        //char c = str.at(1);        //cout<<c;//input abc,  return b        //cout<<str.capacity()<<endl;//return str‘s size,but must be 4n;        //str.clear();        //cout<<str;        //cout<<str.compare("abc");//same as strcmp()        //cout<<str.find_first_of(‘o‘);//find first pos of ‘o‘        //cout<<str.append("abc");//add "abc" after str        //cout<<*str.begin();//same as begin() in vector,so is end()        //int flag = str.find("ab");//return the pos of "ab" in str,-1 will return if not find        //cout<<str.length()<<endl;//return str‘s length,example:"abc"->3        //str.erase(str.begin());//remove the char which is pointed by iterator        //cout<<str.find_last_of(‘o‘)<<endl;//find last pos of char        //str.push_back(‘a‘);//pushback one char        str.push_back(x);        int pos1 = str.find("z");        int pos2 = str.find("j");        int start = 0;        int a = getO(str,start,z);        int b = getO(str,start,j);        int c = getO(str,start,x);       // cout<<a<<b<<c<<endl;               if(a==INT_MIN || b==INT_MIN || c==INT_MIN || b==0)        {            cout<<"Wrong Answer"<<endl;            continue;        }        c-=(b-1)*a;        if(a==c)            cout<<"Accepted"<<endl;        else            cout<<"Wrong Answer"<<endl;    }    return 0;}
题目描述:
对给定的字符串(只包含‘z‘,‘o‘,‘j‘三种字符),判断他是否能AC。

是否AC的规则如下:
1. zoj能AC;
2. 若字符串形式为xzojx,则也能AC,其中x可以是N个‘o‘ 或者为空;
3. 若azbjc 能AC,则azbojac也能AC,其中a,b,c为N个‘o‘或者为空;
输入:
输入包含多组测试用例,每行有一个只包含‘z‘,‘o‘,‘j‘三种字符的字符串,字符串长度小于等于1000。
输出:
对于给定的字符串,如果能AC则请输出字符串“Accepted”,否则请输出“Wrong Answer”。
样例输入:
zojozojoozoojoooozoojoooozoojozojooooozojozojoooo
样例输出:
AcceptedAcceptedAcceptedAcceptedAcceptedAcceptedWrong AnswerWrong Answer

分析:由其生成串的算法可以得出,azbjc为其串的一般式,其中a,b,c表示开始字符到z之间o的数量,b表示字符z到j之间o的数量,c表示字符j到字符串末尾o的数量,且有a=c-(b-1)*a,且b>=1,请注意b=0的情况,应该为wrong answer!

九度OJ 1006 ZOJ