2688-2.cpp

来自「POJ上2688一题多解」· C++ 代码 · 共 18 行

#include <iostream>
#include <cctype>
using namespace std;
int main()
{
	char s[80],*p;
	int c[26]={0};
	gets(s);
	p=s;
	while(*p)
	{
		if (islower(*p)) c[*p-'a']++;
		p++;
	}
	cout<<c['a'-'a']<<' '<<c['e'-'a']<<' '<<c['i'-'a']<<' '<<c['o'-'a']<<' '<<c['u'-'a'];
	return 0;
}