Adventure Time URAL

‘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

5 1

做题的时候注意一下几点

1. 取石块的时候可以不连续

2. 求组合数的时候 注意 c（m，n）=c（m，m-n），不然会爆掉

3. k大于所给的字符串种类的时候，输出全部长度，且种类数为1

4. 当最大的种类小于k的时候，假设加上第二大的数量大于k，注意把整个字符串第二大的数量全部统计完，从中取k-第一大的种类个，进行排列组合运算

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char s[100100];int num[100100];//标记 每一个字符出现的个数-桶排long long int b[100100],bx; //标记每一个种类数出现的个数int cmp(int a, int b){    return a>b;}long long int f(int x, int y) //裸的排列组合计算{    long long int a,b;    a=1;    b=1;    if(y>x/2) //注意这种情况 不然的话，容易爆    {        y=x-y;    }    for(int i=y;i>=1;i--)    {        a=a*i;    }    for(int i=x;i>=x-y+1;i--)    {        b=b*i;    }    return b/a;}int main (){    int n;    while(~scanf("%s %d",s,&n))    {        memset(num,0,sizeof(num));        memset(b,0,sizeof(b));        int len=strlen(s);        for(int i=0; i<len; i++)        {            num[s[i]-'a']++; //统计字符串个数        }        sort(num,num+30,cmp);        long long int Max=0;        for(int i=0;i<n;i++)        {            Max+=num[i]; // 找到最大的max        }        for(int i=0; i<26; i++)        {            b[bx]++;             if(num[i]!=num[i+1])            {                 bx++;                 if(num[i+1]==0)                    break;            }        }       long long  int ans=0;       int flag=0;        for(int i=0;i<bx;i++) // 找到什么情况下满足大于k个了，这个地方注意一定要吧满足大于k个的那个种类的数量全部找到，不要找到大于k的就break。        {            if(ans+b[i]>=n)            {                int x=b[i];                int y=n-ans;                flag=1;                cout<<Max<<' '<<f(x,y)<<endl;                break;            }            else            {                ans+=b[i];            }        }        if(flag==0)        printf("%d 1\n",len);    }    return 0;}