超级抽象的卡时间简单题，快读快写都要加，map映射时要用map<int, int>,用string直接T；
双倍经验： https://www.luogu.com.cn/problem/UVA755

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <math.h>
#include <cstdio>
using namespace std;

int t, n;
map<int, int> cnt;
string ln;

int main()
{
	t = 1;
	while (t)
	{
		cin >> n;
		for (int i = 1; i <= n; i += 1)
		{
			cin >> ln;
			int len = ln.length(), val = 0;
			for (int j = 0; j <= len - 1; j += 1)
			{
				if (ln[j] >= 'A' and ln[j] <= 'Z' or ln[j] >= '0' and ln[j] <= '9')
				{
					val *= 10;
					if (ln[j] == '0') val += 0;
					if (ln[j] == '1') val += 1;
					if (ln[j] >= 'A' && ln[j] <= 'C' || ln[j] == '2') val += 2;
					if (ln[j] >= 'D' && ln[j] <= 'F' || ln[j] == '3') val += 3;
					if (ln[j] >= 'G' && ln[j] <= 'I' || ln[j] == '4') val += 4;
					if (ln[j] >= 'J' && ln[j] <= 'L' || ln[j] == '5') val += 5;
					if (ln[j] >= 'M' && ln[j] <= 'O' || ln[j] == '6') val += 6;
					if (ln[j] >= 'P' && ln[j] <= 'S' || ln[j] == '7') val += 7;
					if (ln[j] >= 'T' && ln[j] <= 'V' || ln[j] == '8') val += 8;
					if (ln[j] >= 'W' && ln[j] <= 'Y' || ln[j] == '9') val += 9;
				}
			}
			cnt[val] += 1;
		}
		bool flag = 0;
		map<int, int>::iterator iter;
		for ( iter = cnt.begin(); iter != cnt.end(); ++iter) if (iter->second >= 2) { printf("%03d-%04d %d\n", iter->first / 10000, iter->first % 10000, iter->second); flag = 1; }
		if (flag == 0) cout << "No duplicates." << endl;
		t -= 1;
	}
}