Codeforces Round #549 (Div. 2) A—C

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原文链接：http://www.cnblogs.com/youchandaisuki/p/10630646.html

本文解析了三道算法竞赛题目：快速关门问题、数字乘积最大化问题及树状结构中的节点删除顺序问题，提供了清晰的思路与高效代码实现。

Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.

There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index

We have to note that Mr. Black opened each door at most once, and in the end all doors became open.

Input

The first line contains integer

The next line contains

It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.

Output

Print the smallest integer

Examples

input

Copy

5
0 0 1 0 0

output

Copy

input

Copy

4
1 0 0 1

output

Copy

Note

In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.

When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.

In the second example when the first two doors were opened, there was open closed door in each of the exit.

With three doors opened Mr. Black was able to use the left exit.

弱智题目记录最后一个0和1的位置输出比较小的那个下标就行

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int INF=1e9+5;

typedef pair<int,int> pii;
int d[maxn],vis[maxn],cnt[maxn],pre[maxn],sum[maxn],val[maxn];
vector<pii> e[maxn];
int n,k;




int main()
{
    cin>>n;
    int a[maxn];
    int x=0,y=0,px,py;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        if(a[i]==0) {x++;px=i;}
        if(a[i]==1) {y++;py=i;}
    }
    if(px>py) cout<<py+1<<endl;
    else cout<<px+1<<endl;
    return 0;
}

View Code

Kurt reaches nirvana when he finds the product of all the digits of some positive integer. Greater value of the product makes the nirvana deeper.

Help Kurt find the maximum possible product of digits among all integers from

Input

The only input line contains the integer

Output

Print the maximum product of digits among all integers from

Examples

input

Copy

output

Copy

input

Copy

output

Copy

input

Copy

1000000000

output

Copy

387420489

Note

In the first example the maximum product is achieved for

In the second example the maximum product is achieved for

In the third example the maximum product is achieved for

给一个数n 输出1-n之内数位乘积最大的乘积。

这个题目最方便的方法就是递归，边递归边比较，可以避免888这样近两位比较出现888优于799的情况。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int INF=1e9+5;

typedef pair<int,int> pii;
int d[maxn],vis[maxn],cnt[maxn],pre[maxn],sum[maxn],val[maxn];
vector<pii> e[maxn];
int n,k;

int cmp(int n)
{
    if(n==0) return 1;
    else if(n<10) return n;
    else
        return max(n%10*cmp(n/10),9*cmp(n/10-1));
}


int main()
{
    char s[15];
    cin>>n;

    cout<<cmp(n)<<endl;
    return 0;
}

View Code

You are given a rooted tree with vertices numerated from

Ancestors of the vertex

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

You noticed that some vertices do not respect others. In particular, if

You decided to delete vertices from the tree one by one. On each step you select such a non-root vertex that it does not respect its parent and none of its children respects it. If there are several such vertices, you select the one with the smallest number. When you delete this vertex

$\text{[math]}$ An example of deletion of the vertex

Once there are no vertices matching the criteria for deletion, you stop the process. Print the order in which you will delete the vertices. Note that this order is unique.

Input

The first line contains a single integer

The next

Output

In case there is at least one vertex to delete, print the only line containing the indices of the vertices you will delete in the order you delete them. Otherwise print a single integer

Examples

input

Copy

output

Copy

1 2 4

input

Copy

output

Copy

-1

input

Copy

output

Copy

Note

The deletion process in the first example is as follows (see the picture below, the vertices with

first you will delete the vertex
the vertex
then you will delete the vertex
the vertex
then you will delete the vertex
you will just delete the vertex
there are no more vertices to delete.

$\text{[math]}$

In the second example you don't need to delete any vertex:

vertices
vertices

$\text{[math]}$

In the third example the tree will change this way:

$\text{[math]}$ 题意就是如果一个点父慈子孝（所有儿子都不respect他，他也不respect他爹）那就把它删掉，从标号最小的点开始删除，比赛的时候弱智了，想着记录所有的儿子，然后看孝子数和儿子数是否一样。现在得得其实只要有一个儿子不是孝子，那就不能删除他。所以反向标记就省事的多，开个数组记录这个点是不是孝子，然后另一个数组记录自己的儿子们有没有真正的孝子，然后判断两个条件是否同时成立就行了，不会打了几十行最后TLE。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int INF=1e9+5;

typedef pair<int,int> pii;
int d[maxn],vis[maxn],cnt[maxn],pre[maxn],sum[maxn],val[maxn];
vector<pii> e[maxn];
int n,k;
int res[maxn];
int c[maxn];

int main()
{
    cin>>n;
    for(int i=0;i<=n;i++) res[i]=1;
    for(int i=1;i<=n;i++)
    {
        int x,y;
        cin>>x>>y;
        if(y==1) c[i]=1;
        if(y==0) res[x]=0;
    }
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        if(res[i]==1 && c[i]==1)
            cout<<i<<" ",cnt++;
    }
    if(cnt==0) cout<<"-1"<<endl;
    return 0;
}

View Code

转载于:https://www.cnblogs.com/youchandaisuki/p/10630646.html