Codeforces Round #499 (Div. 2)--A. Stages
A. Stages
题目链接
Codeforces Round #499 (Div. 2)--A. Stages
题解
简单题。题意大概是给定一组小写字母,每个字母的权重为ai-'a'+1。要求出一个序列满足对于任意ai,ai+1-ai>=2,输出权重之和。
记录每个字母有无出现,直接贪心一遍能取则取即可。
代码
#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int n, k, now, ans, num;
string s;
bool a[233];
int main()
{
cin >> n >> k >> s;
for (auto i : s) {
a[i - 'a' + 1] = true;
}
now = -1;
for (int i = 1; i <= 26; i++) {
if (i > now + 1 && a[i]) {
now = i;
ans += i;
num++;
}
if (num == k)break;
}
if (num == k)cout << ans;
else cout << -1;
}