Hdu 5289 Assignment (to an array, how many intervals to satisfy the difference between the maximum and minimum values in the interval is less than k)

1. The interval is a section, not disconnected yo

2. The code is written in the view of the standard

3. Enumerate the left endpoint, the binary right end-point process:

#include <cstdio> #include <cstring> #include <cmath> #define LL long Long#define Max (a) (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) using namespace Std;const int n=200007;int minn[n][20];//2^18=262144 2^20=1048576int maxx[n][20];//----------- -----------Query O (1)-------------int querymin (int l,int r) {int K=floor (log2 (double) (r-l+1)),//2^k <= (r-l + 1), Flo The OR () takes the whole function down to 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",&AMP;J);        Minn[i][0]=maxx[i][0]=j; }//------------------------------------------pretreatment o (nlogn)---------------for (j=1; (1&LT;&LT;J) <=n; ++J)//1<<j==2^j, enumeration interval length 1,2,4,8,16,,,,, for (I=1; i+ (1<<j) -1<=n; ++i)//i+ (1<<j)-1 denotes interval right boundary, enumeration interval                Left Border {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]); }//-----------------------------------------------------------------------//---------------------------Enumerate the left endpoint,        The two-point right endpoint---------------------------int l,r,mid;            The ans=0;//left endpoint is fixed to I, the right endpoint is determined with L,r,mid, and finally with one of L and R, at this time 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) &L                t;k) {l=mid; } else {r=Mid-1;//himself to demonstrate the algorithm flow know that R can be assigned 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;}

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Hdu 5289 Assignment (to an array, how many intervals to satisfy the difference between the maximum and minimum values in the interval is less than k)