暑假cf50练之1

Codeforces Round #202 (Div. 2)

第一题水题但是wa了一发，排队记录下收到的25，50，100，看能不能找零，要注意100可以找25*3

复杂度O（n）

第二题贪心，先找出最小的花费，然后就能得出最长的位数，然后循环对每个位上的数看能不能加上剩余的油漆比现在这数大，输出即可

复杂度O（10n）

By 2016326603147, contest: Codeforces Round #202 (Div. 2), problem: (B) Color the Fence, Accepted, #
 #include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define MIN(a,b) a<b ? a:b

using namespace std;

const double g=10.0,eps=1e-9;
const int N=100000+10,maxn=10000+10,inf=0x3f3f3f;

int a[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,minn=N,id;
    cin>>n;
    for(int i=1;i<10;i++)
    {
        cin>>a[i];
        if(minn>=a[i])
        {
            minn=a[i];
            id=i;
        }
    }
    if(minn>n)
    {
        cout<<-1<<endl;
        return 0;
    }
    int le=n%minn;
    for(int i=1;i<=(n-n%minn)/minn;i++)
    {
        int maxx=id;
        if(le==0)
        {
            cout<<id;
            continue;
        }
        for(int j=1;j<10;j++)
            if(le+minn>=a[j]&&j>maxx)
               maxx=max(maxx,j);
        cout<<maxx;
        le-=(a[maxx]-minn);
    }
    cout<<endl;
    return 0;
}

b题

第三题推公式，一个多小时还是没做出来

Let the answer be x games. Notice that max(a1,?a2,?…,?an)?≤?x. Then i-th player can be game supervisor in x–ai games. If we sum up we get  — it‘s the number of games in which players are ready to be supervisor. This number must be greater or equal to x — our answer.

Don‘t forget about that condition: max(a1,?a2,?…,?an)?≤?x.

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define MIN(a,b) a<b ? a:b

using namespace std;

const double g=10.0,eps=1e-9;
const int N=100000+10,maxn=10000+10,inf=0x3f3f3f;

ll a[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n,maxx=-1,sum=0;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        sum+=a[i];
        maxx=max(maxx,a[i]);
    }
    ll ans=maxx;
    if(sum%(n-1)==0)cout<<max(sum/(n-1),maxx)<<endl;
    else cout<<max(sum/(n-1)+1,maxx)<<endl;
    return 0;
}

C

D,E现在水平不够，先放着