输入#1
…boyogirlyy…girl…
输出#1
4
2
思路:遍历数组,找到b后遍历其后两个元素是否为oy,如果是则i++
#include<bits/stdc++.h>
using namespace std;
int main()
{
string s;cin >> s;
int len = s.length(),num1=0,num2=0;
int i;
for (i = 0; i < len; i++)
switch (s[i]){
case 'b':
num1++;
if (s[i + 1] == 'o')i++;//遍历b后元素是否为oy,是则下标向后移动
if (s[i + 1] == 'y')i++;
break;
case 'o':
num1++;
if (s[i + 1] == 'y')i++;
break;
case 'y':num1++; break;
case 'g':num2++;
if (s[i + 1] == 'i')i++;
if (s[i + 1] == 'r')i++;
if (s[i + 1] == 'l')i++;
break;
case 'i':num2++;
if (s[i + 1] == 'r')i++;
if (s[i + 1] == 'l')i++;
break;
case 'r':num2++;
if (s[i + 1] == 'l')i++;
break;
case 'l':num2++; break;
default:break;}
cout << num1 << endl << num2;
return 0;
}