计算单词的个数_华为So挑战赛2015年8月份

来源：互联网发布：linux时间戳转换工具编辑：程序博客网时间：2024/05/15 07:38

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#include <iostream>#include <string>  using namespace std;  int Getlen(int *array, int size){int j =0,cnt =0;while(j<size){if (array[j++] >=0){++cnt;}}return cnt;}int main(){string  str;cin>>str;int array1[50];//存放h,u,a所在位置，从0开始int array2[50];int array3[50];for(int i =0; i<50;i++){array1[i] = -1;array2[i]= -1;array3[i] = -1;}int ix = 0;int len = str.size();int fistOfh= str.find_first_of('h',0);int fistOfu= str.find_first_of('u',fistOfh+1);int fistOfa= str.find_first_of('a',fistOfu+1);if (fistOfh == string::npos || fistOfu== string::npos || fistOfa == string::npos){cout<<"0"<<endl;return 0;}else{for (int i = 0; i< len;){int pos = str.find_first_of('h',i);if (pos != string::npos){array1[ix++] = pos;i = pos + 1;}elsebreak;}ix =0;//ix下标重新置0for (int i = fistOfh + 1; i< len;){int pos = str.find_first_of('u',i);if (pos != string::npos){array2[ix++] = pos;i = pos + 1;}elsebreak;}ix =0;//ix下标重新置0for (int i = fistOfh + 1; i< len;){int pos = str.find_first_of('a',i);if (pos != string::npos){array3[ix++] = pos;i = pos + 1;}elsebreak;}}int  cnt1 = Getlen(array1,50);//获得数组中的存放的’h‘’u‘’a‘对应的个数int  cnt2 = Getlen(array2,50);int  cnt3 = Getlen(array3,50);int num =0;for(int i = 0; i< cnt1;i++)for(int j=0; j<cnt2;j++)for (int k = 0; k<cnt3;k++){if (array1[i] < array2[j] && array2[j] < array3[k]){num++;}}cout<<num<<endl;system("pause");return 0;}

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