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Kanade's sum
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1789 Accepted Submission(s): 728
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
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]
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
Source
2017 Multi-University Training Contest - Team 3
自己做的时候想不通 看了别人的题解 怎么人家这么聪明呢qaq
直通车:http://blog.csdn.net/yjf3151731373/article/details/76559550
题意:A 中所有区间内第K大的数求和
分析:遍历A中所有元素 令当前元素为第K大 往右找 上限 再往左找下限 具体看代码qaq
AC:
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
typedef long long LL;
using namespace std;
const int N=1e6;
const LL mod=1e9+7;
int a[N]={0};
int b[N]={0};
int main()
{
int t;scanf("%d",&t);
while(t--)
{
int n,k;scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
LL ans=0;
//当前为第i大;
for(int i=1;i<=n;i++)
{
int x=a[i],num=0,d=0,j;
LL sum=0;
b[++d]=i;//记位置
for(j=i+1;j<=n&&d<k;j++)if(a[j]>x)b[++d]=j;
if(d==k)//右边存a[j]>x使x成为第k大的数
{
num=1; //b当前区间算一个区间
//算后面比它小的数量
for(;j<=n;j++){
if(a[j]<x)num++;
else break;
}
sum+=num;
//算前面比它小的数和比它大的数(前遇到一个比它大的数 后面比它大的数前进一个)
for(j=i-1;j>=1&&d>=1;j--)
{
if(a[j]<x)sum+=num; //因为x是第二大的数区间不能跑到x左边 所以加上x右边的num
else
{
if(d==1)break;//b[1]就是本身 此时在右边界 不能左移了;
num=b[d]-b[d-1]; //右边比它大的数去掉一个区间
d--;
sum+=num;
}
}
}
else{ //右边不存在
for(j=i-1;j>=1&&d<k;j--)if(a[j]>x)b[++d]=j;
if(d==k)
{
sort(b+1,b+d+1);//
num=n-b[d]+1; //n-最大位置+1 = 右边比它大的最后一个位置右边比它小(a[j]<x)的数量;
sum+=num;
for(;j>=1&&b[d]>=i;j--)
{
if(a[j]<x)sum+=num;
else
{
if(b[d]==i)break;// i为右边界
num=abs(b[d]-b[d-1]);
d--;
sum+=num;
}
}
}
}
ans+=x*sum;//sum为区间数
}
printf("%lld\n",ans);
}
return 0;
}