Global Round 2

A - Ilya and a Colorful Walk CodeForces - 1119A

Ilya lives in a beautiful city of Chordalsk.

There are $n$

The houses are colored in colors $c_{1}, c_{2}, \dots, c_{n}$

Ilya wants to select two houses $i$

Ilya loves long walks, so he wants to choose the houses so that the distance between them is the maximum possible.

Help Ilya, find this maximum possible distance.

Input

The first line contains a single integer $n$

The second line contains $n$

It is guaranteed that there is at least one pair of indices $i$

Output

Print a single integer — the maximum possible distance Ilya can walk.

Examples

Input

5
1 2 3 2 3

Output

Input

3
1 2 1

Output

Input

7
1 1 3 1 1 1 1

Output

4
题意：找到两个数，他们数值不相同且距离最远，距离就是两个数位置的下标之差。
解法：贪心的想，要使得距离最远且数值不同，那么就固定一个值在端口，两次遍历取最大值，虽然感觉怪怪的但是就是过了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn=300005;
int n;
int c[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&c[i]);
    int ans=0;
    for(int i=n;i>=1;i--)
    {
        if(c[i]!=c[1])
            ans=max(ans,i-1);
    }
    for(int i=1;i<=n;i++)
    {
        if(c[i]!=c[n])
            ans=max(ans,n-i);
    }
    printf("%d
",ans);
    return 0;
}

B - Alyona and a Narrow Fridge CodeForces - 1119B

Alyona has recently bought a miniature fridge that can be represented as a matrix with $h$

$\text{[math]}$ An example of a fridge with

h = 7

Alyona has $n$

Alyona is interested in the largest integer $k$

Input

The first line contains two integers $n$

The second line contains $n$

Output

Print the single integer $k$

Examples

Input

5 7
2 3 5 4 1

Output

Input

10 10
9 1 1 1 1 1 1 1 1 1

Output

Input

5 10
3 1 4 2 4

Output

5
题意：一个冰箱高h，给你n个瓶子，问最多能放几个瓶子，要求按照题目所给的顺序放瓶子，即[1,2,3,4,5],只能从1开始取，且瓶子只能放在隔板上。
解法：由于是按顺序取瓶子，所以我们可以枚举k，取k个瓶子再排序，再根据贪心，相邻的两个瓶子放在同一层，每一层取较高的那一个，计算高度和，当高度和满足条件时就是我们要求的答案。
注意要开LL， 不然那中间ans 会爆int，导致答案出错。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn=10005;
LL n,h;
int a[maxn],b[maxn];
int main()
{
    scanf("%lld %lld",&n,&h);
    for(int i=0;i<n;i++)
        scanf("%lld",&a[i]);
    for(int k=n;k>=0;k--)
    {
        for(int i=0;i<k;i++)
            b[i]=a[i];
        sort(b,b+k);
        LL ans=0;
        for(int i=k-1;i>=0;i-=2)
            ans+=b[i];
        if(ans<=h)
        {
            printf("%d
",k);
            break;
        }
    }
    return 0;
}

C - Ramesses and Corner Inversion CodeForces - 1119C

Ramesses came to university to algorithms practice, and his professor, who is a fairly known programmer, gave him the following task.

You are given two matrices $A$

$\text{[math]}$ An example of the operation. The chosen submatrix is shown in blue and yellow, its corners are shown in yellow.

Ramesses don't want to perform these operations by himself, so he asks you to answer this question.

A submatrix of matrix $M$

Input

The first line contains two integers $n$

Each of the next $n$

Output

Print "Yes" (without quotes) if it is possible to transform the matrix $A$

Examples

Input

Output

Yes

Input

6 7
0 0 1 1 0 0 1
0 1 0 0 1 0 1
0 0 0 1 0 0 1
1 0 1 0 1 0 0
0 1 0 0 1 0 1
0 1 0 1 0 0 1
1 1 0 1 0 1 1
0 1 1 0 1 0 0
1 1 0 1 0 0 1
1 0 1 0 0 1 0
0 1 1 0 1 0 0
0 1 1 1 1 0 1

Output

Yes

Input

Output

No
题意：给出两个n*m的01矩阵，现在每次可以选择任意一个有四个角的子矩阵（不存在为长度为1的边），并且将四个角取反，问最后是否能将第一个矩阵变为第二个矩阵。
解法：考虑由于每次都取反，所以到最后该操作并不会改变矩阵的整体奇偶性, 进一步思考, 由于在行/列中是两两进行改变, 其实根本不会改变a中某一列/行的奇偶性
这样的话, 如果a中某行/列出现了奇数次的不同, 那么就为No。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn=515;
int n,m;
int cnt;
int a[maxn][maxn],b[maxn][maxn];
int main()
{
    scanf("%d %d",&n,&m);
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        scanf("%d",&a[i][j]);
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        scanf("%d",&b[i][j]);
    for(int i=0;i<n;i++)
    {
        cnt=0;
        for(int j=0;j<m;j++)
        {
            if(a[i][j]!=b[i][j])
                cnt++;
        }
        if(cnt%2!=0)
        {
            printf("No
");
            return 0;
        }
    }
    for(int j=0;j<m;j++)
    {
        cnt=0;
        for(int i=0;i<n;i++)
        {
            if(a[i][j]!=b[i][j])
                cnt++;
        }
        if(cnt%2!=0)
        {
            printf("No
");
            return 0;
        }
    }
    printf("Yes
");
    return 0;
}

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原文地址：https://www.cnblogs.com/jkzr/p/10681276.html