牛客挑战赛34 A~E
闷声发大财
A
O(nmk)dp即可，因为带了1/2的常数+2s所以很稳
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
using namespace std;
int a[801];
int p[801][801];
long long f[801][801];
int n,m,i,j,k,l,K,Y;
int main()
{
// freopen("a.in","r",stdin);
scanf("%d%d%d%d",&n,&m,&K,&Y);
fo(i,1,n)
scanf("%d",&a[i]);
fo(i,1,n)
{
fo(j,1,m)
{
scanf("%d",&p[i][j]);
if (j<Y)
p[i][j]+=a[i]*j;
}
}
memset(f,127,sizeof(f));
f[0][0]=0;
fo(i,0,n-1)
{
fo(j,0,K)
if (f[i][j]<800000000000ll)
{
fd(k,min(K-j,m),0)
f[i+1][j+k]=min(f[i+1][j+k],f[i][j]+p[i+1][k]);
}
}
printf("%lld\n",f[n][K]);
}
B
m=2的约瑟夫问题，给出一个人求n=1~n时剩下这个人的方案&最大的n
约瑟夫的O(n)递推显然过不了，但是随便打表可以发现答案长这样：
1 1 3 1 3 5 7 1 3 5 7 9 11 13 15 1...
随便算
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
using namespace std;
//int f[233];
long long n,m,s,S,ans1,ans2;
int main()
{
// freopen("b.in","r",stdin);
scanf("%lld%lld",&n,&m);
if (!(m&1))
{
printf("0 0\n");
return 0;
}
S=0;s=1;
while (n)
{
if (n>=s)
{
n-=s;
if (m<=s*2-1)
++ans1,ans2=S+(m+1)/2;
}
else
{
if (m<=n*2-1)
++ans1,ans2=S+(m+1)/2;
n=0;
}
S+=s;
s=s*2;
}
printf("%lld %lld\n",ans1,ans2);
return 0;
// f[1]=0;
// fo(i,2,100)
// f[i]=(f[i-1]+2)%i;
//
// fo(i,1,100)
// cout<<f[i]+1<<" ";
// if (!f[i])
// cout<<f[i-1]+1<<" ";
}
C
在普通循环同构的基础上加上了对角互换
懒得画
假设有一条直线穿过圆，并且保证直线上方的数<=对应的直线下方的数
这样解决了对角互换的条件
然后将圆旋转，可以发现上下两部分同时旋转了
于是可以把一对对应点看成一个元素，那么变成元素种数为m(m+1)/2，环的大小为n的普通轮换问题
直接O(n)枚举会挂，所以枚举gcd，再乘上phi(n/gcd)即可
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define mod 19260817
#define Mod 19260815
using namespace std;
int f[mod+1];
int p[1226565];
int T,i,j,k,l,len,s;
long long n,m,ans;
void init()
{
int i,j;
fo(i,1,mod) f[i]=i;
fo(i,2,mod)
{
if (f[i]==i)
{
--f[i];
p[++len]=i;
}
fo(j,1,len)
if ((long long)i*p[j]<=mod)
{
if (!(i%p[j]))
{
f[i*p[j]]=f[i]*p[j];
break;
}
else
f[i*p[j]]=f[i]*(p[j]-1);
}
else
break;
}
}
long long qpower(long long a,int b)
{
long long ans=1;
while (b)
{
if (b&1)
ans=ans*a%mod;
a=a*a%mod;
b>>=1;
}
return ans;
}
int gcd(int n,int m)
{
int r=n%m;
while (r)
{
n=m;
m=r;
r=n%m;
}
return m;
}
int main()
{
// freopen("c.in","r",stdin);
init();
scanf("%d",&T);
for (;T;--T)
{
scanf("%lld%lld",&n,&m);
m=m*(m+1)/2%mod;
s=floor(sqrt(n));
ans=0;
fo(i,1,s)
if (!(n%i))
{
ans=(ans+qpower(m,i)*f[n/i]%mod)%mod;
if (i*i!=n)
ans=(ans+qpower(m,n/i)*f[i]%mod)%mod;
}
// ans=(ans+qpower(m,gcd(n,i)))%mod;
ans=ans*qpower(n,Mod)%mod;
printf("%lld\n",ans);
}
}
D
sb题
把平面旋转45°再扩大\(\sqrt{2}\)倍（即(x,y)-->(x+y,y-x)），变成D*D的矩形操作
排序+扫描线
注意边界不能减
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define low(x) (x&-(x))
#define N 200001
using namespace std;
struct type{
int x,y,s;
} a[200001];
int tr[4*N+1];
int tot,n,D,L,i,j,k,l,x,y,sum;
long long ans;
bool cmp(type a,type b)
{
return a.x<b.x || a.x==b.x && a.s<b.s;
}
void change(int t,int l,int r,int x,int s)
{
int mid=(l+r)/2;
tr[t]+=s;
if (l==r)
return;
if (x<=mid)
change(t*2,l,mid,x,s);
else
change(t*2+1,mid+1,r,x,s);
}
int find(int t,int l,int r,int x,int y)
{
int mid=(l+r)/2,ans=0;
if (x<=l && r<=y)
return tr[t];
if (x<=mid)
ans+=find(t*2,l,mid,x,y);
if (mid<y)
ans+=find(t*2+1,mid+1,r,x,y);
return ans;
}
int main()
{
// freopen("d.in","r",stdin);
scanf("%d%d%d",&n,&D,&L);
fo(i,1,n)
{
scanf("%d%d",&x,&y);
j=x;k=y;
x=j+k;
y=k-j;
a[++tot]={x,y,1};
a[++tot]={x+D,y,-1};
}
sort(a+1,a+tot+1,cmp);
fo(i,1,tot)
{
if (a[i].s==1)
{
ans+=sum-find(1,1,N,max(a[i].y+100001-D+1,1),min(a[i].y+100001+D-1,N));
++sum;
}
change(1,1,N,a[i].y+100001,a[i].s);
}
printf("%lld\n",ans);
}
E
用总数-没有交点的即可，删除可以看成加了一个-1的矩形
求出每个询问上下左右的矩形个数，再减去四个角上的
注意时间也算一维（不用排序），所以需要cdq
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define low(x) (x&-(x))
using namespace std;
struct type{
int x1,y1,x2,y2,s,id;
} a[100001],b[100001];
struct Type{
int s,id;
} c[200001];
struct type1{
int x,s,id;
} d[100001];
struct type2{
int x,y,s,id,Id;
} D[100001];
int tr[200001];
int ans[200001];
int n,i,j,k,l,tp,tot,Tot,sum,len;
bool Cmp(Type a,Type b) {return a.s<b.s;}
bool cmp(type2 a,type2 b) {return a.x<b.x || a.x==b.x && a.id>b.id;}
void change(int t,int s)
{
while (t<=200000)
{
tr[t]+=s;
t+=low(t);
}
}
void clear(int t)
{
while (t<=200000)
{
tr[t]=0;
t+=low(t);
}
}
int find(int t)
{
int ans=0;
while (t)
{
ans+=tr[t];
t-=low(t);
}
return ans;
}
void work1()
{
int i;
fo(i,1,n)
if (d[i].s)
change(d[i].x,d[i].s);
else
ans[d[i].id]-=find(d[i].x-1);
memset(tr,0,sizeof(tr));
}
void work2(int l,int r)
{
int i,mid=(l+r)/2;
if (l==r) return;
work2(l,mid);
work2(mid+1,r);
sort(D+l,D+r+1,cmp);
fo(i,l,r)
if (D[i].Id<=mid && D[i].s)
change(D[i].y,D[i].s);
else
if (D[i].Id>mid && !D[i].s)
ans[D[i].id]+=find(D[i].y-1);
fo(i,l,r)
if (D[i].Id<=mid && D[i].s)
clear(D[i].y);
}
int main()
{
// freopen("e.in","r",stdin);
scanf("%d",&n);
fo(i,1,n)
{
scanf("%d",&tp);
switch (tp)
{
case 1:{
scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
++sum;
a[i].s=1;
b[++tot]=a[i];
break;
}
case 2:{
scanf("%d",&j);
--sum;
a[i]=b[j];
a[i].s=-1;
break;
}
case 3:{
scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
a[i].id=++Tot;
ans[Tot]=sum;
break;
}
}
}
// ---
len=0;
fo(i,1,n)
{
c[++len]={a[i].x1,i};
c[++len]={a[i].x2,-i};
}
sort(c+1,c+len+1,Cmp);
j=0;
fo(i,1,len)
{
j+=i==1 || c[i].s!=c[i-1].s;
if (c[i].id>0)
a[c[i].id].x1=j;
else
a[-c[i].id].x2=j;
}
len=0;
fo(i,1,n)
{
c[++len]={a[i].y1,i};
c[++len]={a[i].y2,-i};
}
sort(c+1,c+len+1,Cmp);
j=0;
fo(i,1,len)
{
j+=i==1 || c[i].s!=c[i-1].s;
if (c[i].id>0)
a[c[i].id].y1=j;
else
a[-c[i].id].y2=j;
}
// ---
fo(i,1,n)
if (a[i].s)
d[i]={a[i].x2,a[i].s,0};
else
d[i]={a[i].x1,0,a[i].id};
work1();
fo(i,1,n)
if (a[i].s)
d[i]={200001-a[i].x1,a[i].s,0};
else
d[i]={200001-a[i].x2,0,a[i].id};
work1();
fo(i,1,n)
if (a[i].s)
d[i]={a[i].y2,a[i].s,0};
else
d[i]={a[i].y1,0,a[i].id};
work1();
fo(i,1,n)
if (a[i].s)
d[i]={200001-a[i].y1,a[i].s,0};
else
d[i]={200001-a[i].y2,0,a[i].id};
work1();
// ---
fo(i,1,n)
if (a[i].s)
D[i]={a[i].x2,a[i].y2,a[i].s,0};
else
D[i]={a[i].x1,a[i].y1,0,a[i].id};
fo(i,1,n)
D[i].Id=i;
work2(1,n);
fo(i,1,n)
if (a[i].s)
D[i]={200001-a[i].x1,200001-a[i].y1,a[i].s,0};
else
D[i]={200001-a[i].x2,200001-a[i].y2,0,a[i].id};
fo(i,1,n)
D[i].Id=i;
work2(1,n);
fo(i,1,n)
if (a[i].s)
D[i]={a[i].x2,200001-a[i].y1,a[i].s,0};
else
D[i]={a[i].x1,200001-a[i].y2,0,a[i].id};
fo(i,1,n)
D[i].Id=i;
work2(1,n);
fo(i,1,n)
if (a[i].s)
D[i]={200001-a[i].x1,a[i].y2,a[i].s,0};
else
D[i]={200001-a[i].x2,a[i].y1,0,a[i].id};
fo(i,1,n)
D[i].Id=i;
work2(1,n);
// ---
fo(i,1,Tot)
printf("%d\n",ans[i]);
}
