5.16pkusc模拟赛1

A.切绳子，不可多条连接，求n条绳子切成k段的最大长度。

卡精度？转成cm二分。n<=10000(5.16)

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define N 10005
using namespace std;
int n,k;
double a
;int b
;
int main(){
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
long long l=1,r=(int)1e7+1;
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++) scanf("%lf",&a[i]),b[i]=(int)((double)a[i]*100+0.5);
while(l<r){
int mid=(l+r)/2;
int ans=0;
for(int i=1;i<=n;i++) ans+=(int)(b[i]/mid);
if(ans<k) r=mid;
else l=mid+1;
}
printf("%.2lf\n",(double)(l-1)/100);
return 0;
}

B.

C.n个数分组，连续的才能分成一组，且每组Max-Min<=k，的方案数。n<=100000(5.16)

rmq+二分/单调队列

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
#define N 100005
#define ll long long
int ffmin
[20],ffmax
[20];
int a
,n,k,T;
ll ans;
bool judge(int from,int to){
//  printf("%d %d\n",from,to);
int l=(int)log2(to-from+1);
int mx=max(ffmax[from][l],ffmax[to-(1<<l)+1][l]);
int mn=min(ffmin[from][l],ffmin[to-(1<<l)+1][l]);
return mx-mn<k;
}
int main(){
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
scanf("%d",&T);
while(T--){
ans=0;
memset(ffmax,0,sizeof(0));
memset(ffmin,0,sizeof(0));
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++) scanf("%d",&a[i]),ffmin[i][0]=a[i],ffmax[i][0]=a[i];
int l=(int)log2(n);
for(int i=1;i<=l;i++)
for(int j=1;j<=n-(1<<i)+1;j++){
ffmin[j][i]=min(ffmin[j][i-1],ffmin[j+(1<<(i-1))][i-1]);
ffmax[j][i]=max(ffmax[j][i-1],ffmax[j+(1<<(i-1))][i-1]);
}
for(int i=1;i<=n;i++){
int l=i,r=n+1;
while(l<r){
int mid=(l+r)/2;
if(judge(i,mid)) l=mid+1;
else r=mid;
}
ans=(ll)(ans+(l-1-i+1));
}
printf("%lld\n",ans);
}
return 0;
}

D.

E.求最小的m，使得[0,m]的线段树存在[l,r]区间节点。

0<=l

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define ll long long
using namespace std;
bool judge(ll l,ll r){
if(l==0) return 1;
if(l<0) return 0;
return judge(l-(r-l+1),r)||judge(l-(r-l+1)-1,r);
}
ll solve(ll l,ll r){
if(r==909){
r++;r--;
}
if(r>2*l) {
//  printf("**%I64d\n",r);
return -1;
}
if(judge(l,r)) return r;
ll len=(r-l+1);
ll ans=solve(l-len-1,r);
ll tmp=solve(l-len,r);
if(ans==-1) ans=tmp;
else if(tmp!=-1) ans=min(ans,tmp);
tmp=solve(l,r+len);
if(ans==-1) ans=tmp;
else if(tmp!=-1) ans=min(ans,tmp);
tmp=solve(l,r+len-1);
if(ans==-1) ans=tmp;
else if(tmp!=-1) ans=min(ans,tmp);
return ans;
}
int main(){
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
ll l,r;
while(~scanf("%I64d%I64d",&l,&r)){
if(l==0) printf("%I64d\n",r);
else printf("%I64d\n",solve(l,r));
}
return 0;
}

F.