Codeforces 514D R2D2 and Droid Army【二分+RMQ】

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An army of n droids is lined up in one row. Each droid is described bym integers a₁, a₂, ..., a_m, wherea_i 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 hasm weapons, the i-th weapon can affect all the droids in the army by destroying one detail of thei-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 mostk 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 containsm integers a₁, a₂, ..., a_m (0 ≤ a_i ≤ 10⁸), where a_i is the number of details of thei-th type for the respective robot.

Output

Print m space-separated integers, where thei-th number is the number of shots from the weapon of thei-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.

Examples

Input

5 2 44 01 22 10 21 3

Output

2 2

Input

3 2 41 21 32 2

Output

1 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.

题目大意：

一共有N个人，每个人有M个属性值，当一个人的所有属性值都小于等于0的时候，这个人就算被销毁了。

我们每次操作可以选一种属性值进行攻击，使得所有人的这个属性的值都-1.

我们最多可以进行K次操作，

问我们最多可以干掉多少个连续的人。

问这种时候的具体操作（每一种属性用了多少次操作）。

思路:

1、比较经典的模型，对于连续的X个人，假如都将其干掉的时候，需要对于每种属性使用的最少操作，就是对应这连续的X个人每种属性的最大值。

2、那么问题转化到区间最大值上来，这里我们可以使用RMQ来解，也可以用线段树来解。

接下来我们可以考虑枚举人数，然后O(NLogN)的去维护当前情况是否可行，直到枚举到不可行为止前的那个答案，就是最终答案。

由此看来，枚举人数是具有单调性的，要干掉更多的人，就需要更多的操作，那么我们可以二分这个人数。

对于可行方案，增加人数，不可行方案，减少人数。

3、二分过程中，维护最后一次可行解的答案，输出即可。

Ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;int maxn[10][200005][20];int a[200060][10];int output[10];int n,m,k;void ST(){    for(int i=1;i<=m;i++)    {        for(int j=1;j<=n;j++)        {            maxn[i][j][0]=a[j][i];        }    }    int len=floor(log10(double(n))/log10(double(2)));    for(int z=1;z<=m;z++)    {        for(int j=1;j<=len;j++)        {            for(int i=1;i<=n+1-(1<<j);i++)            {                maxn[z][i][j]=max(maxn[z][i][j-1],maxn[z][i+(1<<(j-1))][j-1]);            }        }    }}int Slove(int mid){    int ans[10];    for(int i=1;i<=n;i++)    {        if(i+mid-1>n)break;        else        {            int a=i;int b=i+mid-1;            int len= floor(log10(double(b-a+1))/log10(double(2)));            for(int z=1;z<=m;z++)            {                ans[z]=max(maxn[z][a][len], maxn[z][b-(1<<len)+1][len]);            }            int sum=0;            for(int z=1;z<=m;z++)            {                sum+=ans[z];            }            if(sum<=k)            {                for(int z=1;z<=m;z++)                {                    output[z]=ans[z];                }                return 1;            }        }    }    return 0;}int main(){    while(~scanf("%d%d%d",&n,&m,&k))    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                scanf("%d",&a[i][j]);            }        }        ST();        int l=0;        int r=n;        while(r-l>=0)        {            int mid=(l+r)/2;            if(Slove(mid)==1)            {                l=mid+1;            }            else r=mid-1;        }        for(int i=1;i<=m;i++)        {            printf("%d ",output[i]);        }        printf("\n");    }}

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