Codeforces 514 D R2D2 and Droid Army(RMQ+二分法）

An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of thei-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).

A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?

Input

The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.

Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.

Output

Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.

If there are multiple optimal solutions, print any of them.

It is not necessary to make exactly k shots, the number of shots can be less.

Sample test(s)

input
2 4
0
2
1
2
3

output
2

input
2 4
2
3
2

output
3

Note

In the first test the second, third and fourth droids will be destroyed.

In the second test the first and second droids will be destroyed.

让你求最大长度：非常自然想到二分方法,然而还要推断二分到此长度的方案可不可行。

就要找到区间最大值（二维RMQ），RMQ对于查询来说很方便。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int maxn=1e5+10;
int n,m,kk;
int dp[10][maxn][20];//第j种特性
int num[maxn][10];
int res[10],ans[10];
void init()
{
    REPF(i,1,n)
      REPF(j,1,m)//多个
        dp[j][i][0]=num[i][j];
    for(int k=1;k<=m;k++)
      for(int j=1;(1<<j)<=n;j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
              dp[k][i][j]=max(dp[k][i][j-1],dp[k][i+(1<<(j-1))][j-1]);
}
int RMQ(int id,int l,int r)
{
    int k=(int)(log(r-l+1)/log(2.0));
    return max(dp[id][l][k],dp[id][r-(1<<k)+1][k]);
}
bool ok(int x)
{
    for(int i=1;i+x-1<=n;i++)
    {
        int sum=0;
        for(int j=1;j<=m;j++)
        {
            res[j]=RMQ(j,i,i+x-1);
            sum+=res[j];
        }
        if(sum<=kk)
        {
            REPF(j,1,m)  ans[j]=res[j];
            return true;
        }
    }
    return false;
}
void BS()
{
    int l=0,r=n;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(ok(mid)) l=mid+1;
        else r=mid-1;
    }
}
int main()
{
    while(~scanf("%d%d%d",&n,&m,&kk))
    {
        REPF(i,1,n)
          REPF(j,1,m)  scanf("%d",&num[i][j]);
        init();BS();
        REPF(i,1,m)  printf("%d ",ans[i]);
        puts("");
    }
    return 0;
}