Educational Codeforces Round 59 (Rated for Div. 2)
A. Digits Sequence Dividing
题解:因为每个数字都是[1,9]那么直接分成两部分即可 特判n=2的情况
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=3e5+10;
const double eps=1e-8;
#define ll long long
using namespace std;
struct edge{int t,v;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y,int vul){o->t=y;o->v=vul;o->next=h[x];h[x]=o++;}
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
char str[MAXN];
int main(){
int _=read();
while(_--){
int n;n=read();
inc(i,1,n)scanf("%c",&str[i]);
if(n==2&&str[1]>=str[2])printf("NO\n");
else{
puts("YES");
printf("%d\n",2);
printf("%d ",str[1]-'0');
inc(i,2,n)printf("%d",str[i]-'0');
printf("\n");
}
}
}
B. Digital root
题解:打表即可
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 100005
#define nm 1000005
#define pi 3.1415926535897931
const ll inf=998244353;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}
ll n,k;
int main(){
int _=read();while(_--){
n=read();k=read();
printf("%lld\n",9*(n-1)+k);
}
return 0;
}
C. Brutality
题解: 分段考虑 排序取前k个
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=3e5+10;
const double eps=1e-8;
#define ll long long
using namespace std;
struct edge{int t,v;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y,int vul){o->t=y;o->v=vul;o->next=h[x];h[x]=o++;}
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
vector<int>vec;
char str[MAXN];int a[MAXN];
bool cmp(int t1,int t2){return t1>t2;}
int main(){
int n=read();int k=read();
inc(i,1,n)a[i]=read();
scanf("%s",str+1);
vec.pb(a[1]);
ll ans=0;
inc(i,2,n){
if(str[i]==str[i-1])vec.pb(a[i]);
else{
sort(vec.begin(),vec.end(),cmp);
int t1=vec.size();int t2=min(k,t1);
for(int j=0;j<t2;j++)ans+=vec[j];
vec.clear();
vec.pb(a[i]);
}
}
sort(vec.begin(),vec.end(),cmp);
int t1=vec.size();int t2=min(k,t1);
for(int j=0;j<t2;j++)ans+=vec[j];
vec.clear();
printf("%lld\n",ans);
}
D. Compression
题解: bitset暴力莽 是有不需要bitset的做法的
题解:https://www.cnblogs.com/greenty1208/p/10332400.html
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 5205
#define nm 1000005
#define pi 3.1415926535897931
const ll inf=998244353;
using namespace std;
ll read(){
char ch=getchar();
while(!isdigit(ch)&&!('A'<=ch&&ch<='Z'))ch=getchar();
if(isdigit(ch))return ch-'0';
else return ch-'A'+10;
}
bitset<5201>a[NM],tmp;
int n,ans,cnt;
bool check(int t){
int m=n/t-1;
inc(i,0,m)
inc(j,0,m){
int _t=a[i*t+1][j*t+1];
inc(k,2,t)if(a[i*t+1][j*t+k]!=_t)return false;
}
inc(i,0,m){
inc(j,2,t){
tmp=a[i*t+1]^a[i*t+j];
if(tmp.count())return false;
}
}
return true;
}
int main(){
scanf("%d",&n);ans=1;cnt=n;
inc(i,1,n){
inc(j,1,n/4){
int t=read();
inc(k,0,3)if(t&succ(k))a[i].set(4*j-k);
}
}
//inc(i,1,n){inc(j,1,n)printf("%d",(int)a[i][j]);putchar('\n');}
for(int i=2;i*i<=cnt;i++)if(cnt%i==0){
int j=i;
for(;cnt%i==0;j*=i){
cnt/=i;
if(!check(j)){ans*=j/i;break;}
if(cnt%i){ans*=j;break;}
}
while(cnt%i==0)cnt/=i;
}
if(cnt>1)if(check(cnt))ans*=cnt;
printf("%d\n",ans);
}
E. Vasya and Binary String
题解:https://blog.csdn.net/qkoqhh/article/details/86681107
题解:https://www.cnblogs.com/greenty1208/p/10331195.html
F. 待补
G. G. Vasya and Maximum Profit
题解:长得像dp的数据结构题
首先 确定这个表达式 S=a*(r-l+1)-(cl+.....+cr)-max(.....题目的那个式子) (没有md打不出公式)
然后我们对这个式子做个变形 并令k=max(.....)
S=a*r-C[r]-[a*(l-1)-C[l-1]]-k C表示对于c数组求前缀和
令b数组等于 b[i]=a*i-C[i]
让S最大那么b[r]-b[l-1]差值最大
对于常数k我们怎么处理呢 有两种办法 对k最大值分治 然后求跨过这个位置两边的最大最小值 这个可以通过st表/线段树获得
另外一种做法 就是对于每个位置的k的我们单调栈找出他能表示的范围(即最大值就是当前位置的区间) 然后对于每个区间st处理维护 b[r]-b[l-1]的最大值
场上qko大爷的做法 是第一种分治做的 他说他不会写数据结构(滑稽)
吐槽 这个人的st表写的真丑
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 300005
#define nm 1000005
#define pi 3.1415926535897931
const ll inf=998244353;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}
int n;
ll a,ans,d[NM],c[NM],pre[NM],suc[NM],mn[NM],smax[NM][20],smin[NM][20];
struct P{
int i;ll t;
bool operator<(const P&o)const{return t<o.t;}
}st[NM][20];
P query(int l,int r){
int k=mn[r-l];
return max(st[l][k],st[r-succ(k)+1][k]);
}
ll rmq(int l,int r){
int k=mn[r-l];
return max(smax[l][k],smax[r-succ(k)+1][k]);
}
ll _rmq(int l,int r){
int k=mn[r-l];
return min(smin[l][k],smin[r-succ(k)+1][k]);
}
void dfs(int x,int y){
if(x==y){
ans=max(ans,a-c[x]+c[x-1]);
return;
}
P t=query(x,y-1);
ans=max(ans,rmq(t.i+1,y)-_rmq(x-1,t.i)-sqr(t.t));
dfs(x,t.i);dfs(t.i+1,y);
}
int main(){
//freopen("data.in","r",stdin);
n=read();a=read();
inc(i,1,n)d[i]=read(),c[i]=read();
inc(i,1,n)c[i]+=c[i-1];
inc(i,1,n-1)st[i][0]=P{i,d[i+1]-d[i]};
inc(i,1,n)smax[i][0]=smin[i][0]=a*i-c[i];
inc(i,2,n)mn[i]=mn[i/2]+1;
for(int j=1;succ(j)<n;j++)
for(int i=0;i+succ(j)-1<n;i++)
st[i][j]=max(st[i][j-1],st[i+succ(j-1)][j-1]);
for(int j=1;succ(j)<=n;j++)
for(int i=0;i+succ(j)-1<=n;i++)
smax[i][j]=max(smax[i][j-1],smax[i+succ(j-1)][j-1]),
smin[i][j]=min(smin[i][j-1],smin[i+succ(j-1)][j-1]);
dfs(1,n);
return 0*printf("%lld\n",ans);
}
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