Codeforces Round #542 [Alex Lopashev Thanks-Round] (Div. 1)
A - Toy Train
很显然，一个站有多少个糖，那么就要从这个点运多少次。设第i个点有\(a_i\)个糖，那么就要转\(a_i-1\)圈，然后再走一段。很显然最后一段越小越好。
然后枚举起点后，每个点的答案就是起点到他的距离加上再走的距离。然后取个max就好了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cctype>
#define qmin(x,y) (x=min(x,y))
#define qmax(x,y) (x=max(x,y))
#define vi vector<int>
#define vit vector<int>::iterator
#define pir pair<int,int>
#define fr first
#define sc second
#define mp(x,y) make_pair(x,y)
using namespace std;
inline char gc() {
// static char buf[100000],*p1,*p2;
// return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
return getchar();
}
template<class T>
int read(T &ans) {
ans=0;char ch=gc();T f=1;
while(!isdigit(ch)) {
if(ch==EOF) return -1;
if(ch=='-') f=-1;
ch=gc();
}
while(isdigit(ch))
ans=ans*10+ch-'0',ch=gc();
ans*=f;return 1;
}
template<class T1,class T2>
int read(T1 &a,T2 &b) {
return read(a)!=EOF&&read(b)!=EOF?2:EOF;
}
template<class T1,class T2,class T3>
int read(T1 &a,T2 &b,T3 &c) {
return read(a,b)!=EOF&&read(c)!=EOF?3:EOF;
}
typedef long long ll;
const int Maxn=1100000;
const ll inf=0x3f3f3f3f3f3f3f3fll;
int n,m,a[Maxn],b[Maxn],x,y;
signed main() {
// freopen("test.in","r",stdin);
read(n,m);
memset(b,0x3f,sizeof(b));
for(int i=1;i<=m;i++) {
read(x,y);
a[x]++;
if(y>x) qmin(b[x],y-x);
else qmin(b[x],y+n-x);
}
for(int i=1;i<=n;i++) if(!a[i]) b[i]=0;
for(int i=1;i<=n;i++) {
int ans=0;
for(int j=1;j<=n;j++) {
int temp;
if(j>=i) temp=j-i;
else temp=j+n-i;
temp+=(a[j]-1)*n+b[j];
qmax(ans,temp);
}
printf("%d ",ans);
}
return 0;
}
B - Wrong Answer
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cctype>
#define qmin(x,y) (x=min(x,y))
#define qmax(x,y) (x=max(x,y))
#define vi vector<int>
#define vit vector<int>::iterator
#define pir pair<int,int>
#define fr first
#define sc second
#define mp(x,y) make_pair(x,y)
using namespace std;
inline char gc() {
// static char buf[100000],*p1,*p2;
// return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
return getchar();
}
template<class T>
int read(T &ans) {
ans=0;char ch=gc();T f=1;
while(!isdigit(ch)) {
if(ch==EOF) return -1;
if(ch=='-') f=-1;
ch=gc();
}
while(isdigit(ch))
ans=ans*10+ch-'0',ch=gc();
ans*=f;return 1;
}
template<class T1,class T2>
int read(T1 &a,T2 &b) {
return read(a)!=EOF&&read(b)!=EOF?2:EOF;
}
template<class T1,class T2,class T3>
int read(T1 &a,T2 &b,T3 &c) {
return read(a,b)!=EOF&&read(c)!=EOF?3:EOF;
}
typedef long long ll;
const int Maxn=1100000;
const ll inf=0x3f3f3f3f3f3f3f3fll;
int k;
signed main() {
// freopen("test.in","r",stdin);
read(k);
int temp=1999-k%1999;
puts("2000");
for(int i=1;i<=1998;i++) printf("0 ");
printf("%d ",-temp);
printf("%d\n",(k+temp)/1999+temp);
return 0;
}
C - Morse Code
首先，直接n方DP求每一段能代表的字符串的个数很简单，然后因为相同的子串只能统计一次答案，那么就用一颗trie树来存就好了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cctype>
#define qmin(x,y) (x=min(x,y))
#define qmax(x,y) (x=max(x,y))
#define vi vector<int>
#define vit vector<int>::iterator
#define pir pair<int,int>
#define fr first
#define sc second
#define mp(x,y) make_pair(x,y)
using namespace std;
inline char gc() {
// static char buf[100000],*p1,*p2;
// return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
return getchar();
}
template<class T>
int read(T &ans) {
ans=0;char ch=gc();T f=1;
while(!isdigit(ch)) {
if(ch==EOF) return -1;
if(ch=='-') f=-1;
ch=gc();
}
while(isdigit(ch))
ans=ans*10+ch-'0',ch=gc();
ans*=f;return 1;
}
template<class T1,class T2>
int read(T1 &a,T2 &b) {
return read(a)!=EOF&&read(b)!=EOF?2:EOF;
}
template<class T1,class T2,class T3>
int read(T1 &a,T2 &b,T3 &c) {
return read(a,b)!=EOF&&read(c)!=EOF?3:EOF;
}
typedef long long ll;
const int Maxn=11000000;
const int inf=0x3f3f3f3f;
const int mod=1000000007;
int n,f[5100],a[Maxn],ans,ch[Maxn][2],cnt=1;
signed main() {
// freopen("test.in","r",stdin);
read(n);
f[0]=1;
for(int i=1;i<=n;i++)
read(a[i]);
for(int i=1;i<=n;i++) {
memset(f,0,sizeof(f));
f[i+1]=1;
int now=1;
for(int j=i;j>=1;j--) {
for(int k=1;k<=3;k++) f[j]=(f[j]+f[j+k])%mod;
if(j<=i-3) {
if(!a[j]) {
if(a[j+1]) {
if(a[j+2]||!a[j+3]) f[j]=(f[j]+f[j+4])%mod;
}
else {
if(!a[j+2]||!a[j+3]) f[j]=(f[j]+f[j+4])%mod;
}
}
else
if(a[j+3]) {
if(!a[j+1]||!a[j+2]) f[j]=(f[j]+f[j+4])%mod;
}
else if(!a[j+1]||!a[j+2]) f[j]=(f[j]+f[j+4])%mod;
}
if(!ch[now][a[j]]) {
ch[now][a[j]]=++cnt;
ans=(ans+f[j])%mod;
}
now=ch[now][a[j]];
}
printf("%d\n",ans);
}
return 0;
}
D - Isolation
枚举右端点，然后把每个数字最后一次出现的位置设为1，倒数第二次出现的位置设为-1，这样，如果一个后缀和小于等于k，就可以转移。
然后进行分块，每一块记总和和后缀小于等于一个数的dp值的和即可。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cctype>
#include<cmath>
#define qmin(x,y) (x=min(x,y))
#define qmax(x,y) (x=max(x,y))
#define vi vector<int>
#define vit vector<int>::iterator
#define pir pair<int,int>
#define fr first
#define sc second
#define mp(x,y) make_pair(x,y)
using namespace std;
inline char gc() {
// static char buf[100000],*p1,*p2;
// return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
return getchar();
}
template<class T>
int read(T &ans) {
ans=0;char ch=gc();T f=1;
while(!isdigit(ch)) {
if(ch==EOF) return -1;
if(ch=='-') f=-1;
ch=gc();
}
while(isdigit(ch))
ans=ans*10+ch-'0',ch=gc();
ans*=f;return 1;
}
template<class T1,class T2>
int read(T1 &a,T2 &b) {
return read(a)!=EOF&&read(b)!=EOF?2:EOF;
}
template<class T1,class T2,class T3>
int read(T1 &a,T2 &b,T3 &c) {
return read(a,b)!=EOF&&read(c)!=EOF?3:EOF;
}
typedef long long ll;
const int Maxn=110000;
const int inf=0x3f3f3f3f;
const int mod=998244353;
int bn[Maxn],bl[Maxn],br[Maxn],f[Maxn],b[Maxn],a[Maxn],in[Maxn],bf[500][1000],n,k,las[Maxn],pre[Maxn];
inline int modd(int x) {
return x>=mod?x-mod:x;
}
void update(int x) {
memset(bf[x],0,sizeof(bf[0]));
bn[x]=0;
for(int i=br[x];i>=bl[x];i--) {
bf[x][bn[x]+500]=modd(bf[x][bn[x]+500]+f[i]);
bn[x]+=b[i];
}
for(int i=1;i<1000;i++) bf[x][i]=modd(bf[x][i]+bf[x][i-1]);
}
signed main() {
// freopen("test.in","r",stdin);
read(n,k);
int m=sqrt(n);
memset(bl,0x3f,sizeof(bl));
memset(las,-1,sizeof(las));
memset(pre,-1,sizeof(pre));
f[0]=1;
for(int i=0;i<=n;i++) {
in[i]=i/m;
qmin(bl[in[i]],i);
qmax(br[in[i]],i);
}
update(0);
for(int i=1;i<=n;i++) {
read(a[i]);
if(~las[a[i]]) {
pre[i]=las[a[i]];
b[las[a[i]]]=-1;
update(in[las[a[i]]]);
if(~pre[las[a[i]]]) {
b[pre[las[a[i]]]]=0;
update(in[pre[las[a[i]]]]);
}
}
b[i]=1;
int now=1;
for(int j=i-1;j>=bl[in[i]];j--) {
if(now<=k) f[i]=modd(f[i]+f[j]);
now+=b[j];
}
for(int j=in[i]-1;j>=0;j--) {
if(k-now+500>0) f[i]=modd(f[i]+bf[j][min(k-now+500,999)]);
now+=bn[j];
}
update(in[i]);
las[a[i]]=i;
}
printf("%d\n",f[n]);
return 0;
}
