标签：rmq
Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn‘t exceed d.
Note that the order of the points inside the group of three chosen points doesn‘t matter.
The first line contains two integers: n and d (1?≤?n?≤?105; 1?≤?d?≤?109). The next line contains n integers x1,?x2,?...,?xn, their absolute value doesn‘t exceed 109 — the x-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn‘t exceed d.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
4 3 1 2 3 4
4
4 2 -3 -2 -1 0
2
5 19 1 10 20 30 50
1
In the first sample any group of three points meets our conditions.
In the second
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#define inf 2000000000
#define PI acos(-1.0)
#define lson(x) (x<<1)
#define rson(x) ((x<<1)|1)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define drep(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long ll;
int a[100100],n,d;
int minx[100100][30];
int maxx[100100][30];
void init_RMQ(int n)
{
rep(i,1,n)maxx[i][0]=minx[i][0]=a[i];
for(int j = 1; j < 20; ++j)
for(int i = 1; i <= n; ++i)
if(i + (1 << j) - 1 <= n)
{
maxx[i][j] = max(maxx[i][j - 1], maxx[i + (1 << (j - 1))][j - 1]);
minx[i][j] = min(minx[i][j - 1], minx[i + (1 << (j - 1))][j - 1]);
}
}
int getmax(int x,int y){
if(x>y)swap(x,y);
int k=log((y-x+1)*1.0)/log(2.0);
return max(maxx[x][k],maxx[y-(1<<k)+1][k]);
}
int getmin(int x,int y){
if(x>y)swap(x,y);
int k=log((y-x+1)*1.0)/log(2.0);
return min(minx[x][k],minx[y-(1<<k)+1][k]);
}
int main(){
scanf("%d%d",&n,&d);
rep(i,1,n)scanf("%d",&a[i]);
init_RMQ(n);
int l=1;
int r=3;
long long ans=0;
rep(i,3,n){
r=i;
while(getmax(l,r)-getmin(l,r)>d)l++;
if(r-l+1>=3){
//printf("%d %d\n",l,r);
ans+=(r-l)*1LL*(r-l-1)/2LL;
}
}
printf("%I64d\n",ans);
}In the third sample only one group does: {1, 10, 20}.
这题也可以用rmq做，然后依次枚举l,r
codeforces251A. Points on Line
标签：rmq
原文地址：http://blog.csdn.net/kirito_acmer/article/details/45769037
