Educational Codeforces Round 15 套题

这套题最后一题不会，然后先放一下，最后一题应该是大数据结构题

A:求连续最长严格递增的的串，O（n）简单dp

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
typedef  long long LL;
const int N = 1e5+5;
const int INF = 0x3f3f3f3f;
int n,dp[N],a[N],cnt,mx;
int main(){
  scanf("%d",&n);a[0]=INF;
  for(int i=1;i<=n;++i){
    scanf("%d",&a[i]);
    dp[i]=1;
    if(a[i]>a[i-1])dp[i]=dp[i-1]+1;
    mx=max(mx,dp[i]);
  }
  printf("%d
",mx);
  return 0;
}

B:水题，map乱搞

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef  long long LL;
const int N = 1e5+5;
const int INF = 2e9;
map<int,int>mp;
int a[40],cnt;
int main(){ 
  for(int i=0;;++i){
    if((1ll<<i)>INF)break;
    a[++cnt]=(1<<i);
  }
  LL ret=0;
  int x,n;scanf("%d",&n);
  for(int i=1;i<=n;++i){
    scanf("%d",&x);
    for(int j=cnt;j>0;--j){
      if(a[j]<=x)break;
      int tmp=a[j]-x;
      if(mp.find(tmp)!=mp.end())
        ret+=mp[tmp];
    }
    if(mp.find(x)==mp.end())mp[x]=0;
    ++mp[x];
  }
  printf("%I64d
",ret);
  return 0;
}

C:一个典型的二分题，judge如何判断全被覆盖？只要用一下离线求和数组非0就好

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef  long long LL;
const int N = 1e5+5;
const int INF = 2e9;
LL a[N],b[N];
int n,m,c[N];
bool judge(LL r){
  memset(c,0,sizeof(c));
  for(int i=1;i<=m;++i){
    int x=lower_bound(a+1,a+1+n,b[i]-r)-a;
    int y=upper_bound(a+1,a+1+n,b[i]+r)-a;
    ++c[x];--c[y];
  }
  for(int i=1;i<=n;++i){
    c[i]+=c[i-1];
    if(!c[i])return false;
  }
  return true;
}
int main(){
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;++i)
    scanf("%I64d",&a[i]);
  sort(a+1,a+1+n);
  n=unique(a+1,a+1+n)-a-1;
  for(int i=1;i<=m;++i)
    scanf("%I64d",&b[i]);
  sort(b+1,b+1+m);
  m=unique(b+1,b+1+m)-b-1;
  LL l=0,r=INF;
  while(l<r){
    LL mid=(l+r)>>1;
    if(judge(mid))r=mid;
    else l=mid+1;
  }
  printf("%I64d
",(l+r)>>1);
  return 0;
}

D：一个简单的分类讨论，因为最多走k，以k为周期即可

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef  long long LL;
const int N = 1e5+5;
const int INF = 2e9;
LL d,k,a,b,t;
int main(){
  scanf("%I64d%I64d%I64d%I64d%I64d",&d,&k,&a,&b,&t);
  if(d<=k){
    printf("%I64d
",d*a);
    return 0;
  }
  LL ret=k*a;d-=k;
  if(d<=k){
    ret+=min(d*a+t,d*b);
    printf("%I64d
",ret);
    return 0;
  }
  LL t1=k*a+t,t2=k*b;
  if(t2<=t1){
    printf("%I64d
",ret+d*b);
    return 0;
  }
  else {
    ret+=d/k*t1;
    d-=d/k*k;
    if(d==0){printf("%I64d
",ret);return 0;}
    t1=t+d*a,t2=d*b;
    ret+=min(t1,t2);
    printf("%I64d
",ret);
  }
  return 0;
}

E:求从每个点出发路径长度为k的边权和以及边权最小值，刚开始还以为是快速幂，结果发现发现这条路唯一确定，直接倍增即可

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef  long long LL;
const int N = 1e5+5;
const int INF = 2e9;
struct Node{
  int v,mn;
  LL sum;
}f[N][40];
LL retsum[N],k;
int n,retmin[N],cur[N];
void solve(){
  for(int j=0;(1ll<<j)<=k;++j)if(k&(1ll<<j)){
     for(int i=0;i<n;++i){
        retsum[i]+=f[cur[i]][j].sum;
        if(retmin[i]==-1)retmin[i]=f[cur[i]][j].mn;
        else retmin[i]=min(retmin[i],f[cur[i]][j].mn);
        cur[i]=f[cur[i]][j].v;
     } 
  }
  for(int i=0;i<n;++i)
    printf("%I64d %d
",retsum[i],retmin[i]);
}
int main(){
  scanf("%d%I64d",&n,&k);
  for(int i=0;i<n;++i)scanf("%d",&f[i][0].v),cur[i]=i,retmin[i]=-1;
  for(int i=0;i<n;++i)scanf("%d",&f[i][0].mn),f[i][0].sum=f[i][0].mn;
  for(int j=1;(1ll<<j)<=k;++j){
     for(int i=0;i<n;++i){
        f[i][j].v=f[f[i][j-1].v][j-1].v;
        f[i][j].sum=f[i][j-1].sum+f[f[i][j-1].v][j-1].sum;
        f[i][j].mn=min(f[i][j-1].mn,f[f[i][j-1].v][j-1].mn);
     }
  }
  solve();
  return 0;
}

F：不会，看了看别人的代码，并不能看懂，还是太弱