diverta 2019 Programming Contest

　　A：签到。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
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,k;
signed main()
{
	n=read(),k=read();
	cout<<n-k+1;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：签到。背包。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 3010
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 r,g,b,n;
ll f[N];
signed main()
{
	r=read(),g=read(),b=read(),n=read();
	f[0]=1;
	for (int i=r;i<=n;i++) f[i]+=f[i-r];
	for (int i=g;i<=n;i++) f[i]+=f[i-g];
	for (int i=b;i<=n;i++) f[i]+=f[i-b];
	cout<<f[n];
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：太难了吧。先统计一下内部的AB，然后只留下首尾两字符，显然只有BB、AA、BA是有用的。把BA全部接起来，相当于至多剩下一个BA，将其接在BB前或者AA后。AABB两两配对。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 10010
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,ans,f[N][2];
char s[N][12];
signed main()
{
	n=read();
	for (int i=1;i<=n;i++) scanf("%s",s[i]+1);
	for (int i=1;i<=n;i++)
	{
		int m=strlen(s[i]+1);
		for (int j=1;j<m;j++)
		if (s[i][j]=='A'&&s[i][j+1]=='B') ans++;
		if (s[i][1]=='A') f[i][0]=0;
		else if (s[i][1]=='B') f[i][0]=1;
		else f[i][0]=2;
		if (s[i][m]=='A') f[i][1]=0;
		else if (s[i][m]=='B') f[i][1]=1;
		else f[i][1]=2;
	}
	int cnt0=0,cnt1=0,cnt10=0;
	for (int i=1;i<=n;i++)
	{
		if (f[i][0]==1&&f[i][1]==0) cnt10++;
		else if (f[i][1]==0) cnt0++;
		else if (f[i][0]==1) cnt1++;
	}
	if (cnt10) ans+=cnt10-1,cnt10=1;
	if (cnt0&&cnt10) ans++,cnt10=0;
	ans+=min(cnt0+cnt10,cnt1);
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：设n=km+x(0<x<m)。则[n/m]=n%m即k=n-km k(m+1)=n 枚举n的因子即可，注意判断0<x<m。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
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;
}
ll n,ans;
signed main()
{
	cin>>n;
	for (ll i=1;i*i<=n;i++)
	if (n%i==0)
	{
		if ((i-1)*(n/i+1)>n) ans+=i-1;
		if (i*i!=n) if ((n/i-1)*(i+1)>n) ans+=n/i-1;
	}
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　E：做一个异或前缀和，则要选一个x、0交替序列。如果序列异或和不为0，则x只能取该值。否则枚举x。都是要用O(x个数)的复杂度统计序列个数。找出所有x的位置，前缀和求出之间0的个数，然后做一个dp即可，即f[i][0/1]为到第i个位置时最后一个取的是01的方案数，注意一下细节即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 2000010
#define P 1000000007
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,a[N],s[N],id[N],nxt[N],p[N],b[N],f[N][2],m,ans;
int ksm(int a,int k)
{
	int s=1;
	for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
	return s; 
}
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int solve(int x)
{
	if (p[x]<=n)
	{
		id[m=1]=p[x];
		while (nxt[id[m]]<=n) id[m+1]=nxt[id[m]],m++;
	}
	else return 0;
	int t=0;
	for (int i=1;i<=m;i++)
	{
		b[++t]=-1;
		if (i<m) b[++t]=s[id[i+1]]-s[id[i]];
	}
	f[0][0]=1;
	for (int i=1;i<=t;i++)
	{
		f[i][0]=f[i][1]=0;
		if (b[i]==-1)
		{
			f[i][0]=f[i-1][0];
			f[i][1]=(f[i-1][1]+f[i-1][0])%P;
		}
		else
		{
			f[i][1]=f[i-1][1];
			f[i][0]=(f[i-1][0]+1ll*f[i-1][1]*b[i])%P;
		}
	}
	return f[t][1];
}
int solve2(int x)
{
	if (p[x]<=n)
	{
		id[m=1]=p[x];
		while (nxt[id[m]]<n) id[m+1]=nxt[id[m]],m++;
	}
	else return 0;
	int t=0;id[m+1]=n;
	for (int i=1;i<=m;i++)
	{
		b[++t]=-1;
		b[++t]=s[id[i+1]]-s[id[i]];
	}
	f[0][0]=1;
	for (int i=1;i<=t;i++)
	{
		f[i][0]=f[i][1]=0;
		if (b[i]==-1)
		{
			f[i][0]=f[i-1][0];
			f[i][1]=(f[i-1][1]+f[i-1][0])%P;
		}
		else
		{
			f[i][1]=f[i-1][1];
			f[i][0]=(f[i-1][0]+1ll*f[i-1][1]*b[i])%P;
		}
	}
	return f[t][0];
}
signed main()
{
	n=read();
	for (int i=1;i<=n;i++) a[i]=a[i-1]^read();
	for (int i=1;i<=n;i++)
	{
		s[i]=s[i-1];
		if (a[i]==0) s[i]++;
	}
	for (int i=0;i<(1<<20);i++) p[i]=n+1;
	for (int i=n;i>=1;i--)
	{
		nxt[i]=p[a[i]];
		p[a[i]]=i;
	}
	if (a[n]==0)
	{
		ans=ksm(2,s[n]-1);
		for (int i=1;i<(1<<20);i++) inc(ans,solve(i));
	}
	else ans=solve2(a[n]);
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　result：rank 76