hdu 5289 Assignment（给一个数组，求有多少个区间，满足区间内的最大值和最小值之差小于k）

1.区间是一段的，不是断开的哟

2.代码是看着标程写的

3.枚举左端点，二分右端点流程：

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#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;

const int N=200007;
int minn[N][20];//2^18=262144   2^20=1048576
int maxx[N][20];

//----------------------查询O(1)-------------
int queryMin(int l,int r)
{
    int k=floor(log2((double)(r-l+1)));//2^k <= (r - l + 1),floor()向下取整函数
    return Min(minn[l][k],minn[r-(1<<k)+1][k]);
}

int queryMax(int l,int r)
{
    int k=floor(log2((double)(r-l+1)));
    return Max(maxx[l][k],maxx[r-(1<<k)+1][k]);
}
//-------------------------------------------------

int calc(int l,int r)
{
    int k=log2((double)(r-l+1));
    int MAX=Max(maxx[l][k],maxx[r-(1<<k)+1][k]);
    int MIN=Min(minn[l][k],minn[r-(1<<k)+1][k]);
    return MAX-MIN;
}

int main()
{
    int T;
    int n,k,i,j,p;
    LL ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        for(i=1; i<=n; ++i)
        {
            scanf("%d",&j);
            minn[i][0]=maxx[i][0]=j;
        }
//------------------------------------------预处理O(nlogn)---------------
        for(j=1; (1<<j)<=n; ++j)//1<<j==2^j,枚举区间长度1。2。4。8，16。，，。，
            for(i=1; i+(1<<j)-1<=n; ++i)//i+(1<<j)-1表示区间右边界。枚举区间左边界
            {
                p=(1<<(j-1));
                minn[i][j]=Min(minn[i][j-1],minn[i+p][j-1]);
                maxx[i][j]=Max(maxx[i][j-1],maxx[i+p][j-1]);
            }
//-----------------------------------------------------------------------

//---------------------------枚举左端点。二分右端点---------------------------

        int l,r,mid;
        ans=0;
//左端点固定为i，右端点用l,r,mid去确定，最后用l和r中的当中一个。此时l+1==r
        for(i=1; i<=n; ++i)
        {
            l=i,r=n;
            while(l+1<r)
            {
                mid=(l+r)>>1;//(l+r)/2==(l+r)>>1
                if(calc(i,mid)<k)
                {
                    l=mid;
                }
                else
                {
                    r=mid-1;//自己去演示算法流程就知道r能够赋值mid-1
                }
            }
            if(calc(i,r)<k)
            {
                ans=ans+(LL)(r-i+1);
            }
            else
            {
                ans=ans+(LL)(l-i+1);
            }
        }
//---------------------------------------------------------------------------
        printf("%lld\n",ans);
    }
    return 0;
}