《Codeforces Round #681 (Div. 2, based on VK Cup 2019-2020 - Final)》

A：水题：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 2e5+5;
const int M = 1e6+5;
const LL Mod = 998244353;
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

int main()
{
    int ca;ca = read();
    while(ca--)
    {
        int n;n = read();
        for(int i = 4 * n,j = 1;j <= n;i -= 2,++j) printf("%d%c",i,j == n ? '\n' : ' ');
    }
    return 0;
}

B：贪心即可

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 1e5+5;
const int M = 1e6+5;
const LL Mod = 998244353;
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

int main()
{
    int ca;ca = read();
    while(ca--)
    {
        int a,b;a = read(),b = read();
        string s;cin >> s;
        int n = s.size();
        int f = 0,ans = 0,ma = 0;
        for(int i = 0;i < n;++i){
            if(s[i] == '1')
            {
                if(f == 0)
                {
                    if(ans != 0 && b * ma < a) ans += b * ma;
                    else ans += a;
                }
                ma = 0,f = 1;
            }
            else ma++,f = 0;
        }
        printf("%d\n",ans);
    }
    return 0;
}

C：堆维护

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 2e5+5;
const int M = 1e6+5;
const LL Mod = 998244353;
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

struct Node{
    LL x;int id;
    bool operator < (const Node a)const{
    return x < a.x;
    }
};
LL b[N];
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        int n;n = read();
        priority_queue<Node> Q;
        LL sum = 0;
        for(int i = 1;i <= n;++i){
            int x;x = read();
            Q.push(Node{x,i});
        }
        for(int i = 1;i <= n;++i) b[i] = read();
        while(!Q.empty()){
            Node a = Q.top();
            if(sum + b[a.id] <= a.x) sum += b[a.id],Q.pop();
            else break;
        }
        LL ans = 0;
        if(!Q.empty()) ans = Q.top().x;
        ans = max(ans,sum);
        printf("%lld\n",ans);
    }
    return 0;
}

D：没想到这个看起来简单的题能卡我这么久。

主要的思路就是从左边开始删，然后不够的代价从右边补，然后看中间是否某个值 < 0。

亿点细节：一开始是used取最大值，仔细思考之后used应该是叠加的，因为我们每次都让a[i] - used.。

然后我们每次减去之后显然前面的都是前面的区间里的最小值了，所以我们比较的时候应该去比较和最小值的差距。

主要是这里一开始比的都不是和最小值的差距。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 2e5+5;
const int M = 1e6+5;
const LL Mod = 998244353;
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

int a[N];
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        int n;n = read();
        for(int i = 1;i <= n;++i) a[i] = read();
        int f = 0,used = 0,pre = a[1];
        for(int i = 1;i <= n;++i){
            a[i] -= used;
            if(a[i] < 0){f = 1;break;}
            if(a[i] > pre) used += a[i] - pre;
            pre = min(pre,a[i]);
        }
        printf("%s\n",f ? "NO" : "YES");
    }
    system("pause");
    return 0;
}

 E：待补

F：题意是问使答案等于B序列的方案数，一开始没读对。

思路：因为是有顺序的，对于i位置，显然可以删去左边的和删去右边的来得到，这样就有两种可能即方案数 * 2。

然后如果两边只有一边可以删，那就方案数 * 1。

如果两边都要不能删，那就说明不存在满足的方案。

对于能删的情况，我们在遍历中更新即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 2e5+5;
const int M = 1e6+5;
const LL Mod = 998244353;
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline int read()
    {
        int x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
}
using namespace FASTIO;

int a[N],b[N],pos[N];
bool vis[N];
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        memset(vis,0,sizeof(vis));
        int n,k;n = read(),k = read();
        for(int i = 1;i <= n;++i) a[i] = read(),pos[a[i]] = i;
        for(int i = 1;i <= k;++i) b[i] = read(),vis[pos[b[i]]] = 1;
        vis[0] = 1,vis[n + 1] = 1;
        LL ans = 1;
        for(int i = 1;i <= k;++i)
        {
            int ma = pos[b[i]];
            if(vis[ma - 1] == 0 && vis[ma + 1] == 0) ans = ans * 2 % Mod; 
            else if(vis[ma - 1] == 0 || vis[ma + 1] == 0) ans = ans * 1;
            else{ans = 0;break;}
            vis[ma] = 0;
        }
        printf("%lld\n",ans);
    }
    system("pause");
    return 0;
}