2018.08.21【2018提高組】模擬A組 & B組小結
阿新 • • 發佈：2018-11-03
今天看起來有兩個A組，實際上一個也沒有
8:20 看完題目（哇，三道題都好可做啊）
T1打了
做法（stupid++）（雖然
本來就能過）
8:40打完
T2打了個splay（常數爆炸）（stupid++）
9:20打完
T3打80分暴力
9:30打完
（開始無所事事）
10:00 ( ⊙ o ⊙ )啊！T1好像有
做法（碼碼碼）
10:20碼完
10:30 誒Σ(⊙▽⊙”)T2會不會常數太大了（想想想）
10:35 不是可以用二分嗎，
繼續想想想
10:40 先一開始排序一下就好了（碼碼碼）
10:57 碼完（無所事事）
估分100+100+80=280
11:30 比賽結束
估分=得分
T1題解
列舉左上頂點和右上頂點，二維字首和維護
列舉左邊界右邊界，再掃一遍
題目連結http://172.16.0.132/senior/#main/show/5838
T2題解
1：常數十分優美的資料結構加O3
2：二分ans再O(n)判斷
T3題解
還沒切，待填
大概就是DP用樹狀陣列優化一下
T1程式碼來自石家莊的O3
%:pragma GCC optimize(3)
%:pragma GCC optimize("Ofast")
%:pragma GCC optimize("inline")
%:pragma GCC optimize("-fgcse")
%:pragma GCC optimize("-fgcse-lm")
%:pragma GCC optimize("-fipa-sra")
%:pragma GCC optimize("-ftree-pre")
%:pragma GCC optimize("-ftree-vrp")
%:pragma GCC optimize("-fpeephole2")
%:pragma GCC optimize("-ffast-math")
%:pragma GCC optimize("-fsched-spec")
%:pragma GCC optimize("unroll-loops")
%:pragma GCC optimize("-falign-jumps")
%:pragma GCC optimize("-falign-loops")
%:pragma GCC optimize("-falign-labels")
%:pragma GCC optimize("-fdevirtualize")
%:pragma GCC optimize("-fcaller-saves")
%:pragma GCC optimize("-fcrossjumping")
%:pragma GCC optimize("-fthread-jumps")
%:pragma GCC optimize("-funroll-loops")
%:pragma GCC optimize("-fwhole-program")
%:pragma GCC optimize("-freorder-blocks")
%:pragma GCC optimize("-fschedule-insns")
%:pragma GCC optimize("inline-functions")
%:pragma GCC optimize("-ftree-tail-merge")
%:pragma GCC optimize("-fschedule-insns2")
%:pragma GCC optimize("-fstrict-aliasing")
%:pragma GCC optimize("-fstrict-overflow")
%:pragma GCC optimize("-falign-functions")
%:pragma GCC optimize("-fcse-skip-blocks")
%:pragma GCC optimize("-fcse-follow-jumps")
%:pragma GCC optimize("-fsched-interblock")
%:pragma GCC optimize("-fpartial-inlining")
%:pragma GCC optimize("no-stack-protector")
%:pragma GCC optimize("-freorder-functions")
%:pragma GCC optimize("-findirect-inlining")
%:pragma GCC optimize("-fhoist-adjacent-loads")
%:pragma GCC optimize("-frerun-cse-after-loop")
%:pragma GCC optimize("inline-small-functions")
%:pragma GCC optimize("-finline-small-functions")
%:pragma GCC optimize("-ftree-switch-conversion")
%:pragma GCC optimize("-foptimize-sibling-calls")
%:pragma GCC optimize("-fexpensive-optimizations")
%:pragma GCC optimize("-funsafe-loop-optimizations")
%:pragma GCC optimize("inline-functions-called-once")
%:pragma GCC optimize("-fdelete-null-pointer-checks")
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<bits/stdc++.h>
#include <iostream>
#include <cstring>
#include <cstdio>
#define ll long long
using namespace std;
__attribute__((optimize("-O3")))
int n,m;
ll ans;
ll a[201][201],f[201][201];
int main()
{
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
scanf("%lld",&a[i][j]);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
f[i][j]=f[i-1][j]+f[i][j-1]-f[i-1][j-1]+a[i][j];
ans=0;
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
for (int k=i;k<=n;k++)
for (int l=j;l<=m;l++)
ans=max(ans,f[k][l]-f[k][j-1]-f[i-1][l]+f[i-1][j-1]);
printf("%lld",ans);
return 0;
}T1正經程式碼
#include<bits/stdc++.h>
#define fo(i,a,b) for(register int i=a;i<=b;i++)
using namespace std;
long long n,m,a[201][201],qz[201][201],q,ans,sum;
inline void max(long long &x,long long y){x=x>y?x:y;}
int main()
{
scanf("%lld%lld",&n,&m);
fo(i,1,n)fo(j,1,m)scanf("%lld",&a[i][j]);
fo(i,1,n)fo(j,1,m)qz[i][j]=a[i][j]+qz[i][j-1];
fo(i,1,m)fo(j,i,m){sum=0;fo(k,1,n)sum+=(qz[k][j]-qz[k][i-1]),max(ans,sum),sum=(sum<0?0:sum);}
printf("%lld",ans);
}T2各種被卡常的程式碼（來自597大爺）
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
using namespace std;
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 600000
#define LIMIT 1000000000
int n;
long long d;
int ox[MAXN+1],oy[MAXN+1];
int nx[MAXN+1],ny[MAXN+1],maxx,maxy;
int tmp[MAXN+1];
inline bool cmpx(int a,int b){return ox[a]<ox[b];}
inline bool cmpy(int a,int b){return oy[a]<oy[b];}
struct Tree_array{
#define lowbit(x) ((x)&-(x))
int n;
long long t[MAXN+1];
inline void change(int x,long long c){
while (x<=n){
t[x]+=c;
x+=lowbit(x);
}
}
inline long long query(int x){
long long res=0;
while (x){
res+=t[x];
x-=lowbit(x);
}
return res;
}
} numx,sumx,numy,sumy;
int main(){
scanf("%d%lld",&n,&d);
for (int i=1;i<=n;++i)
scanf("%d%d",&ox[i],&oy[i]);
for (int i=1;i<=n;++i)
tmp[i]=i;
sort(tmp+1,tmp+n+1,cmpx);
maxx=0;
for (int i=1;i<=n;++i){
if (ox[tmp[i]]!=ox[tmp[i-1]])
++maxx;
nx[tmp[i]]=maxx;
}
for (int i=1;i<=n;++i)
tmp[i]=i;
sort(tmp+1,tmp+n+1,cmpy);
maxy=0;
for (int i=1;i<=n;++i){
if (oy[tmp[i]]!=oy[tmp[i-1]])
++maxy;
ny[tmp[i]]=maxy;
}
numx.n=sumx.n=maxx;
numy.n=sumy.n=maxy;
for (int i=1;i<=n;++i){
d-=ox[i]*(numx.query(nx[i])*2-numx.query(maxx))-(sumx.query(nx[i])*2-sumx.query(maxx))+ oy[i]*(numy.query(ny[i])*2-numy.query(maxy))-(sumy.query(ny[i])*2-sumy.query(maxy));
if (d<0){
printf("%d\n",i);
return 0;
}
numx.change(nx[i],1);
sumx.change(nx[i],ox[i]);
numy.change(ny[i],1);
sumy.change(ny[i],oy[i]);
}
printf("-1\n");
return 0;
}T2優美二分程式碼
#include<bits/stdc++.h>
using namespace std;
struct ymq{long long x;long long y;};
ymq a[700001],b[700001];
long long c[700001],tot,n,d,i,l,r,mid;
bool cmp(ymq x,ymq y){return x.x<y.x;}
bool check(long long x)
{
tot=0;
long long ans=0,j=0;
for(long long i=1;i<=n;i++)if(a[i].y<=x)tot++,c[tot]=a[i].x;
for(long long i=1;i<=tot;i++)ans=ans+c[i]*(i-1)-j,j+=c[i];
j=0,tot=0;
for(long long i=1;i<=n;i++)if(b[i].y<=x)tot++,c[tot]=b[i].x;
for(long long i=1;i<=tot;i++)ans=ans+c[i]*(i-1)-j,j+=c[i];
if(ans>d)return true;
return false;
}
int main()
{
scanf("%lld%lld",&n,&d);
for(i=1;i<=n;i++)scanf("%lld%lld",&a[i].x,&b[i].x),a[i].y=b[i].y=i;
sort(a+1,a+n+1,cmp);
sort(b+1,b+n+1,cmp);
l=0,r=n+1,mid=(l+r+1)>>1;
while(l<r-1)
{
if(check(mid)==true)
{
r=mid,mid=(l+r+1)>>1;
}
else
{
l=mid,mid=(l+r+1)>>1;
}
}
if(r==n+1)printf("-1");else printf("%lld",r);
}T380分暴力
#include<bits/stdc++.h>
#define mo 1000000007
using namespace std;
long long f[100005],a[100005],qz[100005],i,j,n,m,ans;
long long ksm(long long a,long long b){return(b)?ksm(a*a%mo,b>>1)*((b&1)?a:1)%mo:1;}
int main()
{
scanf("%lld%lld",&n,&m);
for(i=1;i<=n;i++)scanf("%lld",&a[i]),a[i]=(!a[i]?-1:1),qz[i]=qz[i-1]+a[i];
if(n<=8000)
{
f[0]=1;
for(i=1;i<=n;i++)for(j=0;j<=i-1;j++)if(abs(qz[i]-qz[j])<=m)f[i]=f[i]+f[j],f[i]=(f[i]>mo?f[i]-mo:f[i]);
printf("%lld",f[n]);
return 0;
}
if(m==0)
{
for(i=1;i<=n-1;i++)if(qz[i]==0)ans++;
if(qz[n]!=0)
{
printf("0");
return 0;
}
printf("%lld",ksm(2,ans));
return 0;
}
f[0]=1,j=1;
for(i=1;i<=n;i++)
{
f[i]=j,j=j+f[i];
if(i-m>=0)j-=f[i-m];
j=(j+mo)%mo;
}
printf("%lld\n",f[n]);
}