Codeforces Round ＃797 (Div. 3) A~E题解记录

A. Print a Pedestal (Codeforces logo?)

A - Print a Pedestal (Codeforces logo?)

简单找个规律然后按照%3分类讨论就可以了

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define NOTLE ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define endl '\n'
#define T int TT;cin >> TT;while (TT--)
#define lint long long
#define pb push_back
#define eb emplace_back

int main(){
    T{
        int n;
        cin >> n;
        int a1=n/3,m=n%3;
        if(m==0) cout << a1 << ' ' << a1+1 << ' ' << a1-1 <<endl;
        if(m==1) cout << a1 << ' ' << a1+2 << ' ' << a1-1 <<endl;
        if(m==2) cout << a1+1 << ' ' << a1+2 << ' ' << a1-1 <<endl;
    }
    return 0;
}

B. Array Decrements

B - Array Decrements

算出ai-bi的值，然后判断一遍，

WA：存在ai(ai>0)的值不变且其他ai变为0的情况

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define NOTLE ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define endl '\n'
#define T int TT;cin >> TT;while (TT--)
#define lint long long
#define pb push_back
#define eb emplace_back
int a[50010],b[50010],c[50010];
int main(){
    NOTLE;
    T{
        int n,f=1,m=-1,ma=0;
        cin >> n;
        for(int i=1;i<=n;++i) cin >> a[i];
        for(int i=1;i<=n;++i){
            cin >> b[i];
            c[i]=a[i]-b[i];
            if(c[i]<0) f=0;
            if(b[i]==0) ma=max(c[i],ma);
            if(b[i]!=0) m=c[i];
        }
        for(int i=1;i<=n;i++){
            if(c[i]!=m && b[i]!=0) f=0;
            if(b[i]!=0 && c[i]<ma) f=0;
            if(!f) break;
        }
        if(f) cout << "YES" << endl;
        else cout << "NO" <<endl;
    }
    return 0;
}

C. Restoring the Duration of Tasks

C - Restoring the Duration of Tasks

直接判断任务开始时间和上一个任务结束时间的关系就行

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define NOTLE ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define endl '\n'
#define T int TT;cin >> TT;while (TT--)
#define lint long long
#define pb push_back
#define eb emplace_back
int s[500010],f[500010],ans[500010];
int main(){
    NOTLE;
    s[0]=0;
    f[0]=0;
    T{
        int n;
        cin >> n;
        for(int i=1;i<=n;++i) cin >> s[i];
        for(int i=1;i<=n;++i){
            cin >> f[i];
            if(s[i]>=f[i-1]) ans[i]=f[i]-s[i];
            else ans[i]=f[i]-f[i-1];
            cout << ans[i] << ' ';
        }
        cout << endl;
    }
    return 0;
}

D. Black and White Stripe

D - Black and White Stripe

算前缀和来找一段固定长度区间的最大值

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define NOTLE ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define endl '\n'
#define T int TT;cin >> TT;while (TT--)
#define lint long long
#define pb push_back
#define eb emplace_back
int a[500010],d[500010];
int main(){
    NOTLE;
    a[0]=0;
    d[0]=0;
    T{
        int n,k,ans=0;
        cin >> n >> k;
        for(int i=1;i<=n;++i){
            char c;
            cin >> c;
            if(c=='W') a[i]=0;
            else a[i]=1;
            d[i]=d[i-1]+a[i];
            if(i>=k) ans=max(d[i]-d[i-k],ans);
        }
        cout << k-ans << endl;

    }
    return 0;
}

E. Price Maximization

E - Price Maximization

贪心，

WA：在用完之前剩下的小余数和大余数时忘记清空了

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define NOTLE ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define endl '\n'
#define T int TT;cin >> TT;while (TT--)
#define lint long long
#define pb push_back
#define eb emplace_back
lint a[500010],m[1010];
int main(){
    NOTLE;
    T{
        //mas记录大于等于k/2的余数数量，mis则记录剩下的小余数数量
        lint n,k,ans=0,mas=0,mis=0;
        cin >> n >> k;
        memset(m,0,sizeof(m));
        for(int i=1;i<=n;++i){
            cin >> a[i];
            ++m[a[i]%k];
            ans+=a[i]/k;
        }

        for(int i=k/2;i>=1;i--){
            if(k-i==i) ans+=m[i]/2,mas+=m[i]%2;
            else{
                if(m[i]>=m[k-i]){
                    ans+=min(m[i],m[k-i]);
                    mis+=m[i]-m[k-i];
                }
                if(m[i]<m[k-i]){
                    ans+=min(m[i]+mas+mis,m[k-i]);
                    if(m[i]+mis>=m[k-i]) mis=m[i]+mis-m[k-i];
                    else if(m[i]+mis+mas>=m[k-i]) mas=m[i]+mis+mas-m[k-i],mis=0;
                    else mas=m[k-i]-(m[i]+mis+mas),mis=0;
                }
            }
        }
        ans+=mas/2;
        cout << ans << endl;
    }
    return 0;
}