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
