Adventure Time URAL - 2024

原创于 2018-08-11 21:20:51 发布 · 291 阅读

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在一个危险的洞穴中，主角芬恩与影之守护者战斗，同时杰克需要收集黑暗之石并确保它们的颜色种类不超过限定数量以避免唤醒宇宙邪恶力量。本文通过一个故事背景介绍了一种算法问题，即如何在限定不同颜色石头数量的情况下，找到最大数量的安全石头集合及其组合数量。

‘Here’s for you!’ shouted Finn fighting the Shadow Guardians. ‘Jake! Collect the Darkness Rocks and let’s run from here!’

Jake would have been glad to finish as soon as possible, because it was wet and sad in the Dangerous Cave, but it was not so easy to gather Rocks. So he called BMO and asked him to help with choosing an optimal set of Rocks. Jake told BMO that there were n Rocks in the Cave, and they were numbered with different integers from 1 to n. Each Rock was painted one of 26 colors. Jake also reminded BMO of Princess Bubblegum’s warning: there should be at most k Rocks of pairwise different colors among the Rocks taken from the Cave. Only a set of Rocks with this property is safe. If one takes an unsafe set of Rocks from the Cave, Cosmic Evil will be awakened! Of course, Jake doesn’t want to awaken the Evil, but he wants to take as many Darkness Rocks from the Cave as possible. Help BMO to find the size of the largest safe set of Rocks and the number of different safe sets of this size. Two sets of Rocks are different if one of them contains a Rock with a number that is not contained in the other set.

Input

The first line contains n lowercase English letters (1 ≤ n ≤ 10 5); each letter denotes the color of a Rock. In the second line you are given the integer k (1 ≤ k≤ 26).

Output

Output two integers separated with a space: the size of the largest safe set of Darkness Rocks and the number of different safe sets of this size.

Example

input	output
abcde 1	1 5
ababac 2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
using namespace std;
typedef long long ll;
char a[100010];
ll x[533];
long long cc[105];
bool cmp(ll a,ll b)
{
    return a>b;
}
ll C(ll m,ll n)
{
    long long i;
    cc[0] = 1;
    for(i = 1; i <= m; i++)cc[i] = cc[i - 1] * (n - i + 1) / i;
    return cc[m];
}
ll vis[130];
int main()
{
    //cout << C(1,26)<<endl;
    ll n,i,j,w,q,e,sum, len;
    scanf("%s",a);
    len = strlen(a);
    scanf("%lld",&n);
    memset(vis,0,sizeof(vis));
    memset(x,0,sizeof(x));
    sum = 0;
    for(i=0;i<len;i++)
    {
        x[a[i]-'a']++;
        if(vis[a[i]] == 0) {vis[a[i]] = 1;sum++;}
    }
//    cout << sum << endl;
//    getchar();
    if(sum <= n) {
        printf("%lld 1\n",len);
        return 0;
    }
    sort(x,x+26,cmp);
    ll num = 0;
    for(i=0;i<n;i++)
    {
        num += x[i];
    }
    w = x[n-1];
    q = 0;e = 0;
    //cout << w << endl;
    for(i=0;i<26;i++)
    {
        if(x[i]==w) q++;
        if(x[i]>w) e++;
    }
    //cout << n - e << " " << q << endl;
    //getchar();
    ll ans = C(n-e,q);
    printf("%lld %lld\n",num, ans);
    return 0;
}