标签：art 遇到 order rom NPU -- 怎么 博客 tps
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 64655 | Accepted: 30135 | |
| Case Time Limit: 2000MS | ||
Description
For the daily milking, Farmer John‘s N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.
Input
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
Source
#include<iostream>
#include<string.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mod 1000000007
#define INF 0x3f3f3f3f
#define MAX 50005
int Max[MAX],Min[MAX];
int a[MAX];
int n,q;
inline void read(int &x){
char ch;
bool flag=false;
for (ch=getchar();!isdigit(ch);ch=getchar())if (ch==‘-‘) flag=true;
for (x=0;isdigit(ch);x=x*10+ch-‘0‘,ch=getchar());
x=flag?-x:x;
}
inline void write(int x){
static const int maxlen=100;
static char s[maxlen];
if (x<0) { putchar(‘-‘); x=-x;}
if(!x){ putchar(‘0‘); return; }
int len=0; for(;x;x/=10) s[len++]=x % 10+‘0‘;
for(int i=len-1;i>=0;--i) putchar(s[i]);
}
int lowbit(int x)
{
return x&-x;
}
void updata(int i,int val)
{
while(i<=n)
{
Min[i]=min(Min[i],val);
Max[i]=max(Max[i],val);
i+=lowbit(i);
}
}
int query(int l,int r)
{
int maxn=a[l],minn=a[r];
while(1)
{
maxn=max(maxn,a[r]),minn=min(minn,a[r]);
if(l==r) break;
for(r-=1;r-l>=lowbit(r);r-=lowbit(r))
maxn=max(Max[r],maxn),minn=min(minn,Min[r]);
}
return maxn-minn;
}
int main()
{
memset(Min,INF,sizeof(Min));
read(n);read(q);
for(int i=1;i<=n;i++){
read(a[i]);
updata(i,a[i]);
}
while(q--)
{
int a,b;
read(a),read(b);
write(query(a,b));
putchar(‘\n‘);
}
return 0;
}
时间复杂度近似为log2(n)。
标签：art 遇到 order rom NPU -- 怎么 博客 tps
原文地址：https://www.cnblogs.com/zhgyki/p/9545316.html
