Educational Codeforces Round 63 Div. 2

　　A：找到两个相邻字符使后者小于前者即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 300010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
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;
}
int n;char s[N];
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read();scanf("%s",s+1);
	int x=0;
	for (int i=2;i<=n;i++) if (s[i]<s[i-1]) {x=i;break;}
	if (x==0) cout<<"NO";
	else cout<<"YES"<<endl<<x-1<<' '<<x;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：设(n-11)/2=x，8的出现次数=y。显然x>=y时，后手一直拿8就可以阻止先手取胜。考虑x<y的情况，后手的最优策略仍然是一直拿最靠前的8，而先手应该拿最靠前的非8的数。于是找到第x+1个8的位置check一下。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
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;
}
int n;char s[N];
signed main()
{
	n=read();scanf("%s",s+1);
	int cnt=0;for (int i=1;i<=n;i++) if (s[i]=='8') cnt++;
	if ((n-11)/2>=cnt) {cout<<"NO";return 0;}
	cnt=(n-11)/2;
	for (int i=1;i<=n;i++)
	if (s[i]=='8')
	{
		cnt--;
		if (cnt==-1)
		{
			if (i-1<=n-11) cout<<"YES";
			else cout<<"NO";
			break;
		}
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：a相邻作差取gcd，看b中是否有gcd的因子。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 300010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
ll gcd(ll n,ll m){return m==0?n:gcd(m,n%m);}
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;
}
int n,m;
ll a[N],b[N];
signed main()
{
	n=read(),m=read();
	for (int i=1;i<=n;i++) a[i]=read();
	for (int i=1;i<=m;i++) b[i]=read();
	sort(a+1,a+n+1);
	ll u=0;
	for (int i=2;i<=n;i++) u=gcd(u,a[i]-a[i-1]);
	for (int i=1;i<=m;i++)
	if (u%b[i]==0)
	{
		cout<<"YES"<<endl;
		cout<<a[1]<<' '<<i;
		return 0;
	}
	cout<<"NO";
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：显然修改的子串一定会被选入答案，这样序列可以被分为五部分，即不被选入答案→被选入答案但不修改→被选入答案且修改→被选入答案但不修改→不被选入答案。可以整一些前缀和，但写起来可能有点麻烦。更简单的做法是f[i][0/1/2/3/4]表示当前在哪个阶段。https://www.cnblogs.com/Gloid/p/10358542.html受这一场的D启发。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 300010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
ll gcd(ll n,ll m){return m==0?n:gcd(m,n%m);}
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;
}
int n,m,a[N];
ll ans,s[N],ssuf[N],pre[N],suf[N],f[N][5];
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read(),m=read();
	for (int i=1;i<=n;i++) a[i]=read();
	memset(f,200,sizeof(f));
	for (int j=0;j<5;j++) f[0][j]=0;
	for (int i=1;i<=n+1;i++)
	{
		for (int j=0;j<5;j++)
			for (int k=0;k<=j;k++)
			f[i][j]=max(f[i][j],f[i-1][k]);
		f[i][1]+=a[i];
		f[i][2]+=1ll*a[i]*m;
		f[i][3]+=a[i];
	}
	cout<<f[n+1][4];
}

　　E：写着脸上的拉格朗日插值。直接代入求值会T，先求出多项式即可。或者也可以高斯消元。所以询问次数是用来干啥的？

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define P 1000003
#define N 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
ll gcd(ll n,ll m){return m==0?n:gcd(m,n%m);}
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;
}
int a[20],Inv[P],f[20],g[20];
int inv(int a){return Inv[a];}
/*int calc(int n,int x)
{
    int ans=0;
    for (int i=0;i<n;i++)
    {
        int u=1;for (int j=0;j<n;j++) if (i!=j) u=1ll*u*(P+i-j)%P;
        u=1ll*a[i]*inv(u)%P;
        for (int j=0;j<n;j++) if (i!=j) u=1ll*u*(P+x-j)%P;
        ans=(ans+u)%P;
    }
    return ans;
}*/
int calc(int x)
{
	int s=f[11];
	for (int i=10;i>=0;i--) s=(1ll*s*x+f[i])%P;
	return s;
}
signed main()
{
	Inv[0]=Inv[1]=1;for (int i=2;i<P;i++) Inv[i]=P-1ll*(P/i)*Inv[P%i]%P;
	for (int i=0;i<12;i++)
	{
		cout<<"?"<<' '<<i<<endl;
		a[i]=read();
	}
	for (int i=0;i<12;i++)
	{
		memset(g,0,sizeof(g));g[0]=1;
		for (int j=0;j<12;j++)
		if (i!=j)
		{
			for (int k=0;k<12;k++) g[k]=1ll*g[k]*inv((i-j+P)%P)%P;
			for (int k=11;k>=1;k--) g[k]=(g[k-1]-1ll*g[k]*j%P+P)%P;
			g[0]=(P-1ll*g[0]*j%P)%P;
		}
		for (int j=0;j<12;j++) f[j]=(f[j]+1ll*g[j]*a[i])%P;
	}
	for (int i=0;i<P;i++)
	if (calc(i)==0) {cout<<"!"<<' '<<i<<endl;return 0;}
	cout<<"!"<<' '<<-1<<endl;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　F：怎么感觉这么经典！哇我好像做过原题！https://www.cnblogs.com/Gloid/p/10291212.html 这个题一方面去掉了边权且没有重边，一方面要输出方案。前者当然是喜大普奔，不用判一些乱七八糟的情况了。后者求出dp数组后递归输出即可，在预处理能否串成链时记录一下串成的链是怎样的。复杂度O(2ⁿ*n³+3ⁿ*n²)，但这个dp的常数一看就优秀的没边。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 14
#define M 1000
#define inf 100000000
#define rep(i,t,S) for (int t=S,i=lg2[t&-t];t;t^=t&-t,i=lg2[t&-t])
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
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;
}
int n,m,a[N][N],f[1<<N],size[1<<N],lg2[1<<N];
bool g[N][N][1<<N];
int way[N][N][1<<N][N+1];
struct data{int x,y,z;
}edge[M];
void print(int S)
{
	if (size[S]==1) return;
	for (int i=S-1&S;i;i=i-1&S)
    rep(x,u,i^S) rep(y,v,i^S)
    {
        if (g[x][y][i]&&f[i^S]+size[i]+1==f[S])
		{
			cout<<y+1<<' '<<way[x][y][i][1]+1<<endl;
			for (int t=1;t<way[x][y][i][0];t++) cout<<way[x][y][i][t]+1<<' '<<way[x][y][i][t+1]+1<<endl;
			cout<<way[x][y][i][way[x][y][i][0]]+1<<' '<<x+1<<endl;
			print(i^S);
			return;
		}
        if (x==y) break;
    }
}
int main()
{
    n=read(),m=read();
    for (int i=1;i<=m;i++)
	{
		edge[i].x=read()-1,edge[i].y=read()-1,edge[i].z=1;
		g[edge[i].x][edge[i].y][0]=g[edge[i].y][edge[i].x][0]=1;
	}
    memset(f,42,sizeof(f));for (int i=0;i<n;i++) f[1<<i]=0,lg2[1<<i]=i;
	for (int i=1;i<(1<<n);i++) size[i]=size[i^(i&-i)]+1;
    for (int i=1;i<(1<<n);i++)
    {
        rep(x,p,(1<<n)-1^i) rep(y,q,(1<<n)-1^i)
        if (x!=y||i!=(i&-i)) rep(j,o,i)
			if (g[j][y][i^(1<<j)]&&g[x][j][0])
			{
				g[x][y][i]=1;
				for (int u=0;u<=way[j][y][i^(1<<j)][0];u++) way[x][y][i][u]=way[j][y][i^(1<<j)][u];
				way[x][y][i][++way[x][y][i][0]]=j;
				break;
			}
    }
    for (int i=1;i<(1<<n);i++)
        for (int j=i-1&i;j;j=j-1&i)
            rep(x,u,i^j) rep(y,v,i^j)
            {
                if (g[x][y][j]&&f[i^j]+size[j]+1<f[i]) f[i]=f[i^j]+size[j]+1;
                if (x==y) break;
            }
    printf("%d\n",f[(1<<n)-1]);
    print((1<<n)-1);
    return 0;
}

　　小小号打的。result：rank 1 rating +297 精准的没有超过小号（