POJ 1002

思路：很坑爹的一道水题么，题目没说字符串有多长，一开始开的长度是30,一直RE。另外如果没有重复的输出的是No duplicates.,注意有句点。。。两种解法：map或者二叉搜索树中序遍历一次，维护一个cnt（记录次数）域。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
using namespace std;
char str[300];
string temp;
map<string, int>ss;
int a[26] = {2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5,
    6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9
};
void Switch(){
    char s[300];
    memset(s, 0, sizeof(str));
    temp.clear();
    int k = 0;
    int len = strlen(str);
    for(int i = 0;i < len;i ++){
        if(str[i] >= '0' && str[i] <= '9') s[k++] = str[i];
        if(str[i] >= 'A' && str[i] <= 'Z') s[k++] = a[str[i]-'A'] + '0';
        if(k == 3) s[k++] = '-'; 
    }
    temp = s;
}
int main(){
    int n;
    /* freopen("in.c", "r", stdin); */
    while(~scanf("%d", &n)){
        ss.clear();
        for(int i = 0; i < n;i ++){
            memset(str, 0, sizeof(str));
            scanf("%s", str);
            Switch();
            ss[temp]++;
        }
        int flag = 0;
        for(map<string, int>::iterator it = ss.begin(); it != ss.end(); ++it){
            if(it->second > 1){
                flag = 1;
                cout << it->first << " " << it->second << endl;
            }
        }
        if(flag == 0)
            printf("No duplicates.
");
    }
    return 0;
}