codeforces624B Making a String 贪心排序

最新推荐文章于 2016-08-29 14:51:49 发布

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本文介绍了一个算法问题：如何构造一个由特定字母组成的最长字符串，同时满足每个字母出现次数的限制及不同字母出现次数各不相同的条件。通过排序和标记已使用次数的方法，实现了高效的求解。

You are given an alphabet consisting of n letters, your task is to make a string of the maximum possible length so that the following conditions are satisfied:

the i-th letter occurs in the string no more than ai times;
the number of occurrences of each letter in the string must be distinct for all the letters that occurred in the string at least once.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 26) — the number of letters in the alphabet.

The next line contains n integers ai (1 ≤ ai ≤ 109) — i-th of these integers gives the limitation on the number of occurrences of the i-th character in the string.

Output

Print a single integer — the maximum length of the string that meets all the requirements.

Examples

input
3
2 5 5

output
11

input
3
1 1 2

output
3

Note

For convenience let's consider an alphabet consisting of three letters: "a", "b", "c". In the first sample, some of the optimal strings are: "cccaabbccbb", "aabcbcbcbcb". In the second sample some of the optimal strings are: "acc", "cbc".

从大到小排序，用一个map来唯一标示，遇到标记过的不断减一，直到减到这个数之前没有出现过，累加，依次进行到底

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <map>

using namespace std;

int a[30];
map<int, int> m;

bool cmp(int x, int y) {
	return x > y;
}

int main()
{
	int n;
	while (~scanf("%d", &n)) {
		m.clear();
		for (int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
		}
		sort(a, a + n, cmp);
		long long ans = 0;
		for (int i = 0; i < n; i++) {
			if (m[a[i]] == 0) {
				m[a[i]]++;
				ans += a[i];
			} else {
				while (a[i] > 0) {
					if (m[--a[i]] == 0) {
						m[a[i]]++;
						ans += a[i];
						break;
					}
				}
			}
		}
		printf("%I64d\n", ans);
	}
	return 0;
}