HZOI20190822模拟29题解
题面:https://www.cnblogs.com/Juve/articles/11396238.html
下面开始一句话题解:
A:爬山:
二分答案,check即可
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define int long long
using namespace std;
const int mod=1e9+7;
int n,d,a,b,l,r,mid,ans;
bool check(int m){
int p=abs(m-a),q=abs(m-b),x,y;
if(p%d==0) x=p/d;
else x=p/d+1;
if(q%d==0) y=q/d;
else y=q/d+1;
if(x+y<=n-1) return 1;
return 0;
}
signed main(){
scanf("%lld%lld%lld%lld",&n,&d,&a,&b);
if(d==0){
printf("%lld\n",max(a,b));
return 0;
}
l=min(a,b)-n*d,r=max(a,b)+n*d;
ans=max(a,b);
while(l<=r){
mid=(l+r)>>1;
if(check(mid))
ans=max(ans,mid),l=mid+1;
else r=mid-1;
}
printf("%lld\n",ans);
return 0;
}
B:学数数:
离散化,单调栈,前缀和
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#define int long long
using namespace std;
const int MAXN=1e5+5;
int n,q,a[MAXN],b[MAXN],k,cnt,sum[MAXN];
int sta[MAXN],top=0,l[MAXN],r[MAXN];
char op[5];
signed main(){
scanf("%lld%lld",&n,&q);
for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
memcpy(b,a,sizeof(a));
sort(b+1,b+n+1);
cnt=unique(b+1,b+n+1)-b-1;
for(int i=1;i<=n;i++){
while(top!=0&&a[sta[top]]<=a[i]) top--;
if(top==0) l[i]=1;
else l[i]=sta[top]+1;
sta[++top]=i;
}
top=0;
for(int i=n;i>=1;i--){
while(top!=0&&a[sta[top]]<a[i]) top--;
if(top==0) r[i]=n;
else r[i]=sta[top]-1;
sta[++top]=i;
}
for(int i=1;i<=n;i++){
int pos=lower_bound(b+1,b+cnt+1,a[i])-b;
sum[pos]+=(i-l[i]+1)*(r[i]-i+1);
}
for(int i=2;i<=n;i++) sum[i]+=sum[i-1];
while(q--){
scanf("%s%lld",op,&k);
if(op[0]=='='){
int x=upper_bound(b+1,b+cnt+1,k)-b-1;
int y=lower_bound(b+1,b+cnt+1,k)-b;
if(x==y) printf("%lld\n",sum[x]-sum[x-1]);
else puts("0");
}else if(op[0]=='<'){
int x=lower_bound(b+1,b+cnt+1,k)-b-1;
printf("%lld\n",sum[x]);
}else{
int x=upper_bound(b+1,b+cnt+1,k)-b-1;
printf("%lld\n",sum[cnt]-sum[x]);
}
}
return 0;
}
C:七十和十七
推一个公式即可,加上快速幂
#include<cstdio>
#include<iostream>
#include<algorithm>
#define int long long
using namespace std;
const int MAXN=1e5+5;
const int mod=1e9+7;
int f[MAXN],n,bit[MAXN];
int q_pow(int a,int b,int p){
int res=1;
while(b){
if(b&1) res=1ll*res*a%p;
a=1ll*a*a%p;
b>>=1;
}
return res%mod;
}
signed main(){
scanf("%lld",&n);
bit[0]=1;
for(int i=1;i<=n;i++) bit[i]=(bit[i-1]*2)%mod;
for(int i=1;i<=n;i++)
f[i]=(f[i-1]+(bit[i-1]-1)%mod*q_pow(i,mod-2,mod)%mod)%mod;
printf("%lld\n",f[n]);
return 0;
}

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