Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 368 Accepted Submission(s): 243
解析:说了那么多,其实很简单,就是求有多少种字母。
#include <cstdio>
#include <cstring>
const int MAXN = 1e5+5;
char s[MAXN];
bool vis[30];
int main()
{
int t, cn = 0;
scanf("%d", &t);
while(t--){
scanf("%s", s);
memset(vis, 0, sizeof(vis));
int res = 0;
for(int i = 0; s[i] != ‘\0‘; ++i){
if(!vis[s[i]-‘a‘]){
vis[s[i]-‘a‘] = true;
++res;
}
}
printf("Case #%d: %d\n", ++cn, res);
}
return 0;
}
原文:http://www.cnblogs.com/inmoonlight/p/5773467.html