Codeforces Round #499 (Div. 2) A，B

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本文解析了两道竞赛编程题目，A题介绍了字符处理与排序的方法，通过特定算法解决字符组合问题；B题则关注资源分配策略，利用二分查找确定最优天数方案，确保团队成员在限定条件下获取充足食物。

A题：感觉做麻烦了。。。。

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,k;
    cin>>n>>k;
    char s[55];
    int a[55];
    cin>>s;
    for(int i=0; i<n; i++)
    {
        a[i]=s[i]-96;
    }
    sort(a,a+n);
    int sum=0;
    int total=0;
    int flag=0;
    for(int i=0; i<n; i++)
    {
        if(total<k&&a[i]!=a[i+1])
        {
            if(a[i]+1==a[i+1]&&flag==0)
            {
                flag=1;
                sum+=a[i];
                total++;
                continue;
            }
            if((a[i]+1!=a[i+1])&&flag==0)
            {
                sum+=a[i];
                total++;
                continue;
            }
            else
                flag=0;
        }
    }
    if(total==k)
        cout<<sum<<endl;
    else
        cout<<"-1"<<endl;
}

B：
Planning The Expedition
Natasha is planning an expedition to Mars for nn people. One of the important tasks is to provide food for each participant.
The warehouse has mm daily food packages. Each package has some food type ai.
Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.
Formally, for each participant jj Natasha should select his food type bjbj and each day jj-th participant will eat one food package of type bj. The values bjbj for different participants may be different.
What is the maximum possible number of days the expedition can last, following the requirements above?
Input
The first line contains two integers nn and mm (1≤n≤100) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers a1,a2,…,am (1≤ai≤100), where aiai is the type of ii-th food package.
Output
Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0.
input
4 10
1 5 2 1 1 1 2 5 7 2
output
2
input
100 1
1
output
0
input
2 5
5 4 3 2 1
output
1
input
3 9
42 42 42 42 42 42 42 42 42
output
3
做法：
用sum+=每种类的个数/天数，如果sum>=人数，可以，然后二分找最大。

#include<bits/stdc++.h>
using namespace std;
vector<int>vec;
int n;
int cmp(int a,int b)
{
    return a>b;
}
bool check(int x)
{
    int sum=0;
    if(x==0)
        return true;
    for(int i=0;i<vec.size();i++)
    {
        sum+=vec[i]/x;
    }
    if(sum>=n)
        return true;
    else
        return false;
}
int main()
{
    int m;
    cin>>n>>m;
    int cnt[111];
    memset(cnt,0,sizeof(cnt));
    while(m--)
    {
        int x;
        scanf("%d",&x);
        cnt[x]++;
    }
    for(int i=1;i<=100;i++)
    {
        if(cnt[i])
            vec.push_back(cnt[i]);
    }
    sort(vec.begin(),vec.end(),cmp);
    int l=0,r=*vec.begin();
    int ans=0;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(check(mid))
        {
            l=mid+1;
            ans=mid;
        }
        else
            r=mid-1;
    }
    cout<<ans<<endl;
}