hdu 6058 Kanade's sum
2017-08-02 12:42
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Problem Description
Give you an array A[1..n]of
length n.
Let f(l,r,k) be
the k-th largest element of A[l..r].
Specially , f(l,r,k)=0 if r−l+1<k.
Give you k ,
you need to calculate ∑nl=1∑nr=lf(l,r,k)
There are T test cases.
1≤T≤10
k≤min(n,80)
A[1..n] is a permutation of [1..n]
∑n≤5∗105
Input
There is only one integer T on first line.
For each test case,there are only two integers n,k on
first line,and the second line consists of n integers
which means the array A[1..n]
Output
For each test case,output an integer, which means the answer.
Sample Input
1
5 2
1 2 3 4 5
Sample Output
30
题目分析:第一眼看求区间第K大,主席树。。。结果发现要求的区间居然有平方个。。。
实际上就是一道模拟题,枚举每一个x,对于x,找到x之前的K个比x大的点,和x之后比x大的点。那么就可以在这个2k的区间里面找到所有x是第K大的区间。具体细节详见代码。
代码入下:
Give you an array A[1..n]of
length n.
Let f(l,r,k) be
the k-th largest element of A[l..r].
Specially , f(l,r,k)=0 if r−l+1<k.
Give you k ,
you need to calculate ∑nl=1∑nr=lf(l,r,k)
There are T test cases.
1≤T≤10
k≤min(n,80)
A[1..n] is a permutation of [1..n]
∑n≤5∗105
Input
There is only one integer T on first line.
For each test case,there are only two integers n,k on
first line,and the second line consists of n integers
which means the array A[1..n]
Output
For each test case,output an integer, which means the answer.
Sample Input
1
5 2
1 2 3 4 5
Sample Output
30
题目分析:第一眼看求区间第K大,主席树。。。结果发现要求的区间居然有平方个。。。
实际上就是一道模拟题,枚举每一个x,对于x,找到x之前的K个比x大的点,和x之后比x大的点。那么就可以在这个2k的区间里面找到所有x是第K大的区间。具体细节详见代码。
代码入下:
/* Author:kzl */ #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; typedef long long LL; const int maxx = 5e5+10; int t,n,k,a[maxx]; int l[maxx],r[maxx]; int main(){ scanf("%d",&t); while(t--){ scanf("%d%d",&n,&k); for(int i=0;i<n;i++)scanf("%d",&a[i]); LL ans=0;l[0] = r[0] = 0; for(int i=0;i<n;i++) { int lcnt=1,rcnt=1,j; for( j=i+1;j<n;j++) { if(rcnt>k)break; if(a[j]>a[i])r[rcnt++]=j-i;//r存储的是距离当前点的距离 } if(j>=n)r[rcnt]=n-i;//在i右边没有K个比a[i]大的数的情况下。i右边最后比a[i]大的数的右边还有比i小的一些数,存储下来。 for(j=i-1;j>=0;j--) { if(lcnt>k)break; if(a[j]>a[i])l[lcnt++]=i-j; } if(j<=0) l[lcnt]=i+1; for(j=0;j<lcnt;j++) { if(k-j-1>=rcnt)continue; int lnum=l[j+1]-l[j];//就是i右边第j+1个比a[i]大的数与第j个比a[i]大的数之间有多少个数。 int rnum=r[k-j]-r[k-j-1];//因为两边区间内加起来比a[i]大的数要是k-1个。 ans+=(LL)a[i]*lnum*rnum; } } printf("%lld\n",ans); } return 0; }
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