20190814校内模拟赛
T1.坏掉的键盘 (keyboard)
传送门
Description
\(n\times m\)网格，从左上走到右下，只能往下或往右
有一些障碍物，数量\(\leq 10^5\)
询问能否到达，\(n,m\leq 10^9\)
Solution
直接暴力维护每行可以到达的区间即可
Code
//70pts 离散后暴力算方案数
#include<bits/stdc++.h>
#define ll long long
#define db double
#define reg register
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
#define int ll
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return x*f;
}
const int MN=4e5+5,P=19260817;
int Mul(int x,int y){return (1ll*x*y)%P;}
int Add(int x,int y){return (x+y)%P;}
int nn,mm,n,m,k,x[MN],y[MN];
int num[MN],tot,fac[MN],inv[MN];
bool mp[6005][6005];int f[6000][6005];
int lb(int _){return lower_bound(num+1,num+tot+1,_)-num;}
bool chk()
{
if(k==0) return true;
tot=0;reg int i,j;
for(i=1;i<=k;++i)
{
num[++tot]=x[i];
if(x[i]>1) num[++tot]=x[i]-1;
if(x[i]<n) num[++tot]=x[i]+1;
}
num[++tot]=1;num[++tot]=n;
std::sort(num+1,num+tot+1);
tot=unique(num+1,num+tot+1)-num-1;
for(i=1;i<=k;++i) x[i]=lb(x[i]);n=lb(n);
tot=0;
for(i=1;i<=k;++i)
{
num[++tot]=y[i];
if(y[i]>1) num[++tot]=y[i]-1;
if(y[i]<m) num[++tot]=y[i]+1;
}
num[++tot]=1;num[++tot]=m;
std::sort(num+1,num+tot+1);
tot=unique(num+1,num+tot+1)-num-1;
for(i=1;i<=k;++i) y[i]=lb(y[i]);m=lb(m);
memset(mp,0,sizeof mp);
for(i=1;i<=k;++i) mp[x[i]][y[i]]=true;
f[1][1]=1;
for(i=1;i<=n;++i)for(j=1;j<=m;++j)if(i!=1||j!=1)
{
if(!mp[i][j])f[i][j]=Add(f[i-1][j],f[i][j-1]);
else f[i][j]=0;
}
return f[n][m]>0;
}
signed main()
{
#ifndef LOCAL
freopen("keyboard.in","r",stdin);
freopen("keyboard.out","w",stdout);
#endif
int T=read();
while(T--)
{
nn=n=read(),m=mm=read(),k=read();
reg int i;
for(i=1;i<=k;++i) x[i]=read(),y[i]=read();
printf("%lld\n",chk()?(ll)nn+(ll)mm-2ll:-1ll);
}
return 0;
}
//100 pts
#include<bits/stdc++.h>
#define ll long long
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
using namespace std;
#define reg register
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
return x*f;
}
const int MN=1e5+5;
struct node{int x,y;bool operator<(const node&o)const{return x<o.x||(x==o.x&&y<o.y);}};
struct sec{int x,l,r;};
sec P[MN<<2];node p[MN];int K,k;
void mk(int x,int l,int r){if(x==1&&l!=1)return;P[++K]=(sec){x,l,r};}
int n,m;
int main()
{
#ifndef LOCAL
freopen("keyboard.in","r",stdin);
freopen("keyboard.out","w",stdout);
#endif
int T=read();
while(T--)
{
n=read(),m=read(),k=read();K=0;
reg int i,j;p[0]=(node){1,0};p[k+1]=(node){n,m+1};
for(i=1;i<=k;++i) p[i].x=read(),p[i].y=read();
std::sort(p+1,p+k+1);
for(j=1,i=1;i<=k+1;++i)
{
if(p[i].x==p[i-1].x)
{
if(p[i].y>1+p[i-1].y)
mk(j,p[i-1].y+1,p[i].y-1);
}
else
{
if(p[i-1].y<m) mk(j,p[i-1].y+1,m);
if(p[i-1].x+1<p[i].x) mk(++j,1,m);
++j;if(p[i].y>1) mk(j,1,p[i].y-1);
}
}
int N=j;
for(i=1,j=2;j<=K;++j)
{
if(P[j].l>P[j].r){P[j].l=m+1;continue;}
while(P[i].x<P[j].x-1&&i<=K){++i;while(P[i].l>P[i].r&&i<=K)++i;}
while(P[i].x<P[j].x&&P[i].r<P[j].l&&i<=K){++i;while(P[i].l>P[i].r&&i<=K)++i;}
if(i<j&&P[i].x==P[j].x-1&&P[i].r>=P[j].l) P[j].l=max(P[j].l,P[i].l);
else P[j].l=m+1;if(P[j].l>P[j].r) P[j].l=m+1;
}
//for(i=1;i<=K;++i)dbg1(P[i].x),dbg1(P[i].l),dbg2(P[i].r);
printf("%lld\n",(P[K].x==N&&P[K].l<=m&&P[K].r==m)?((ll)n+(ll)m-2ll):-1ll);
}
return 0;
}
T2. LibreOJ (libreoj)
传送门
Description
给定\(b\)，多组询问
每个询问给出\(n\)，求出\(\sum_{i=1}^n if(i)\)
其中\(f(x)=atoi(sort(itoa(x,b)),b)\)
\(itoa(x,b)\)表示把\(x\)转为\(b\)进制的字符串
\(sort(s)\)表示将字符串\(s\)按照字典序升序排序后的字符串
\(atoi(s,b)\)表示将字符串\(s\)以\(b\)今之形式转换得到的正整数
对\(998244353\)取模
Solution
\(11223446\) 产生了 \(11111111\)（大于等于 \(1\) 的）加\(111111\) （大于等于\(2\)的）加 \(1111\)（大于等于 \(3\)的）加\(111\) （大于等于 \(4\) 的）加 \(1\)（大于等于 \(5\) 的）加 \(1\)（大于等于 \(6\) 的）加 \(0\)（大于等于\(7,8,9\)的）的贡献。
然后发现仍然是可以数位dp的，就是枚举\(1-9\)，算有\(i\)个数位大于等于它的数的一次和（我还算了零次和，便于转移）是多少，把之前的\(b=2\)的暴力改改就行了。。。
Code
//50 pts dp(b=2)+暴力
#include<bits/stdc++.h>
#define ll long long
#define db double
#define reg register
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return x*f;
}
const int P=998244353;
int Mul(int x,int y){return 1ll*x*y%P;}
int Add(int x,int y){return (x+y)%P;}
int T,b,len[25];
char n[25][1005];
namespace solve1
{
int f[100005];
int to_int(char *s,int L)
{
int r=0;
for(int i=0;i<L;++i) r=r*b+(s[i]-'0');
return r;
}
int cal(int i)
{
char tmp[40];int L=-1;
while(i) tmp[++L]='0'+(i%b),i/=b;++L;
std::sort(tmp,tmp+L);
return to_int(tmp,L);
}
void Main()
{
for(int i=1;i<=100000;++i) f[i]=Add(f[i-1],Mul(i,cal(i)));
for(int i=1;i<=T;++i) printf("%d\n",f[to_int(n[i],len[i])]);
}
}
namespace solve2
{
char s[1005];
int L,g[1005][1005],f[1005][1005],g2[1005][1005],f2[1005][1005];
void Main()
{
for(int o=1;o<=T;++o)
{
reg int i,j;
scanf("%s",s);L=strlen(s);
memset(g,0,sizeof g);
memset(f,0,sizeof f);
memset(g2,0,sizeof g2);
memset(f2,0,sizeof f2);
f[L-1][0]=f[L-1][1]=g[L-1][0]=1;
f2[L-1][1]=1;
if(s[L-1]=='1') g[L-1][1]=1,g2[L-1][1]=1;
int _=1;
for(i=L-2;~i;--i)
{
_=Mul(_,2);
f[i][0]=f[i+1][0];
for(j=1;j<=L-i;++j) f[i][j]=Add(f[i+1][j-1],f[i+1][j]);
for(j=1;j<=L-i;++j) f2[i][j]=Add(Add(f2[i+1][j-1],Mul(_,f[i+1][j-1])),f2[i+1][j]);
g[i][0]=g[i+1][0];
for(j=1;j<=L-i;++j)
{
if(s[i]=='1') g[i][j]=Add(g[i+1][j-1],f[i+1][j]),
g2[i][j]=Add(Add(g2[i+1][j-1],Mul(_,g[i+1][j-1])),f2[i+1][j]);
else g[i][j]=g[i+1][j],g2[i][j]=g2[i+1][j];
}
}
int nm=1,ans=0;
for(i=1;i<=L;++i,nm=Add(1,Mul(nm,2))) ans=Add(ans,Mul(nm,g2[0][i]));
printf("%d\n",ans);
}
}
}
int main()
{
#ifndef LOCAL
freopen("libreoj.in","r",stdin);
freopen("libreoj.out","w",stdout);
#endif
T=read(),b=read();int ml=0,i;
if(b==2) {solve2::Main();return 0;}
for(i=1;i<=T;++i) scanf("%s",n[i]),ml=max(ml,len[i]=strlen(n[i]));
if(len[i]<=40) solve1::Main();
return 0;
}
#include<bits/stdc++.h>
#define ll long long
#define db double
#define reg register
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return x*f;
}
const int P=998244353;
int Mul(int x,int y){return 1ll*x*y%P;}
int Add(int x,int y){return (x+y)%P;}
int T,b;
namespace solve3
{
char s[1005];
int L,g[1005][1005],f[1005][1005],g2[1005][1005],f2[1005][1005];
int nm[1005],ans;
int cal(int l,int r){return (l+r)*(r-l+1)/2;}
void Main()
{
reg int i,j,_;
for(_=1,j=1;j<=1000;++j,_=Add(1,Mul(_,b)))nm[j]=_;
for(int o=1;o<=T;++o)
{
ans=0;
scanf("%s",s);L=strlen(s);
for(int d=1;d<b;++d)
{
memset(g,0,sizeof g);memset(f,0,sizeof f);memset(g2,0,sizeof g2);memset(f2,0,sizeof f2);
f[L-1][0]=d;f2[L-1][0]=cal(0,d-1);g[L-1][0]=min(s[L-1]-'0',d-1)+1;g2[L-1][0]=cal(0,min(s[L-1]-'0',d-1));
f[L-1][1]=b-d;f2[L-1][1]=cal(d,b-1);if(s[L-1]-'0'>=d) g[L-1][1]=s[L-1]-'0'+1-d,g2[L-1][1]=cal(d,s[L-1]-'0');
for(_=1,i=L-2;~i;--i)
{
_=Mul(_,b);f[i][0]=Mul(f[i+1][0],d);f2[i][0]=Add(Mul(d,f2[i+1][0]),Mul(cal(0,d-1),Mul(_,f[i+1][0])));
for(j=1;j<=L-i;++j) f[i][j]=Add(Mul(b-d,f[i+1][j-1]),Mul(d,f[i+1][j]));
for(j=1;j<=L-i;++j) f2[i][j]=Add(Add(Mul(b-d,f2[i+1][j-1]),Mul(cal(d,b-1),Mul(_,f[i+1][j-1]))),
Add(Mul(d,f2[i+1][j]),Mul(cal(0,d-1),Mul(_,f[i+1][j]))));
if(s[i]-'0'>=d)
{
g[i][0]=Mul(d,f[i+1][0]);
g2[i][0]=Add(Mul(d,f2[i+1][0]),Mul(cal(0,d-1),Mul(_,f[i+1][0])));
}
else
{
g[i][0]=g[i+1][0],g2[i][0]=Add(g2[i+1][0],Mul(s[i]-'0',Mul(_,g[i+1][0])));
if(s[i]>'0') g[i][0]=Add(g[i][0],Mul(f[i+1][0],s[i]-'1'+1)),
g2[i][0]=Add(g2[i][0],Add(Mul(s[i]-'0',f2[i+1][0]),Mul(cal(0,s[i]-'1'),Mul(_,f[i+1][0]))));
}
for(j=1;j<=L-i;++j)
{
if(s[i]-'0'>=d)
{
g[i][j]=Add(g[i+1][j-1],Mul(d,f[i+1][j]));
g2[i][j]=Add(Add(g2[i+1][j-1],Mul(s[i]-'0',Mul(_,g[i+1][j-1]))),
Add(Mul(d,f2[i+1][j]),Mul(cal(0,d-1),Mul(_,f[i+1][j]))));
if(s[i]-'1'>=d) g[i][j]=Add(g[i][j],Mul(s[i]-'1'+1-d,f[i+1][j-1])),
g2[i][j]=Add(g2[i][j],Add(Mul(s[i]-'0'-d,f2[i+1][j-1]),Mul(cal(d,s[i]-'1'),Mul(_,f[i+1][j-1]))));
}
else
{
g[i][j]=g[i+1][j],g2[i][j]=Add(g2[i+1][j],Mul(s[i]-'0',Mul(_,g[i+1][j])));
if(s[i]>'0') g[i][j]=Add(g[i][j],Mul(f[i+1][j],s[i]-'1'+1)),
g2[i][j]=Add(g2[i][j],Add(Mul(f2[i+1][j],s[i]-'0'),Mul(cal(0,s[i]-'1'),Mul(_,f[i+1][j]))));
}
}
}
for(i=1;i<=L;++i)ans=Add(ans,Mul(g2[0][i],nm[i]));
}
printf("%d\n",ans);
}
}
}
int main()
{
#ifndef LOCAL
freopen("libreoj.in","r",stdin);
freopen("libreoj.out","w",stdout);
#endif
T=read(),b=read();int ml=0,i;
solve3::Main();
return 0;
}
T3.一键挖矿 (veinminer)
传送门
Description
求一个矩阵（其中的数是个排列）中有多少的子矩阵\(S\)满足\(max_S-min_S=siz_S\)
也就是矩阵中的数恰好是连续的一段数
Solution
这里有坑
Code
//大暴力 30pts
#include<bits/stdc++.h>
#define ll long long
#define db double
#define reg register
#define dbg1(x) cerr<<#x<<"="<<" "
#define dbg2(x) cerr<<#x<<"="<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return x*f;
}
int n,m;
ll ans;
namespace solve1
{
int a[2005];
void Main()
{
reg int i,j,mi,ma;
for(i=1;i<=m;++i) a[i]=read();
for(i=1;i<=m;++i)
for(mi=ma=a[i],j=i;j<=m;++j)
{
mi=min(mi,a[j]);
ma=max(ma,a[j]);
if(ma-mi+1==j-i+1) ++ans;
}
}
}
namespace solve2
{
int a[105][105],Mi[105][105][105],Ma[105][105][105];
void Main()
{
reg int i,j,k,l,mi,ma;
for(i=1;i<=n;++i)for(j=1;j<=m;++j)a[i][j]=read();
for(l=1;l<=n;++l)
{
for(i=1;i<=m;++i)
for(mi=ma=a[l][i],j=i;j<=m;++j)
{
mi=min(mi,a[l][j]);
ma=max(ma,a[l][j]);
Mi[l][i][j]=mi;
Ma[l][i][j]=ma;
}
}
for(l=1;l<=n;++l)for(k=1;k<=m;++k)
for(j=k;j<=m;++j)for(ma=0,mi=10001,i=l;i<=n;++i)
{
mi=min(mi,Mi[i][k][j]);
ma=max(ma,Ma[i][k][j]);
if(ma-mi+1==(i-l+1)*(j-k+1)) ++ans;
}
}
}
int main()
{
#ifndef LOCAL
freopen("veinminer.in","r",stdin);
freopen("veinminer.out","w",stdout);
#endif
n=read(),m=read();
if(n==1&&m<=2000) solve1::Main();
else if(n<=100&&m<=100) solve2::Main();
else
{
for(int i=1;i<=n*m;++i) read();
ll lin=1ll*n*1ll*m*1ll*(m+1)/2;
ll row=1ll*n*1ll*(n-1)/2;
ans=lin+row;
}
return 0*printf("%lld\n",ans);
}
gugugu
gugugugugugu
gugugugugugugugugu
gugugugugugugugugugugugu
......
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