Codeforces Round #292 (Div. 2)

最新推荐文章于 2025-08-01 11:12:48 发布

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最新推荐文章于 2025-08-01 11:12:48 发布

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Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).

Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.

Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?

Input

You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.

Output

If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).

Otherwise, print "Yes".

Sample test(s)

input
5 5 11

output
No

input
10 15 25

output
Yes

input
0 5 1

output
No

input
0 0 2

output
Yes

Note

In fourth sample case one possible route is: $\text{[math]}$ .

题意：在一张无限大的坐标上，起点在（0，0），只能上下左右走，可以原路返回，给定终点（a,b)，给定走的步数s，问能否刚好s步到达终点。

思路：如果（0,0）到（a，b）的最少步数径大于给定的s，那么怎么都不可能。如果小于等于，只要判断s-最少步数是否是2的倍数。因为你可以在两个地点来回走，每一个来回增加2步。增加的肯定是二的倍数。比如，你从（0,0）到（1,0），可以一步到达，也可以（0,0）->(0,-1)->(0,0)->(1,0),

绕弯路的情况也类似。

#include<iostream>
using namespace std;
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#define maxn 205
#define inf 0x3f3f3f3f
int main()
{
    int a,b,s;
    while(cin>>a>>b>>s){
        if(s<abs(a)+abs(b))cout<<"No"<<endl;
        else{
            if((s-(abs(a)+abs(b)))%2==0)cout<<"Yes"<<endl;
            else cout<<"No"<<endl;
        }
    }
}

Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.

There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $\text{[math]}$ -th boy and $\text{[math]}$ -th girl to have dinner together (as Drazil is programmer, i starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever.

Drazil wants to know whether he can use this plan to make all his friends become happy at some moment.

Input

The first line contains two integer n and m (1 ≤ n, m ≤ 100).

The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.

The third line conatins integer g (0 ≤ g ≤ m), denoting the number of happy girls among friends of Drazil, and then follow g distinct integers y1, y2, ... , yg (0 ≤ yj < m), denoting the list of indices of happy girls.

It is guaranteed that there is at least one person that is unhappy among his friends.

Output

If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No".

Sample test(s)

input
2 3
0
1 0

output
Yes

input
2 4
1 0
1 2

output
No

input
2 3
1 0
1 1

output
Yes

Note

By $\text{[math]}$ we define the remainder of integer division of i by k.

In first sample case:

On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes at this day.
On the 2-nd day, Drazil invites 0-th boy and 2-nd girl. Because 0-th boy is already happy he makes 2-nd girl become happy at this day.
On the 3-rd day, Drazil invites 1-st boy and 0-th girl. 0-th girl is happy, so she makes 1-st boy happy.
On the 4-th day, Drazil invites 0-th boy and 1-st girl. 0-th boy is happy, so he makes the 1-st girl happy. So, all friends become happy at this moment.

思路：简单模拟，坑点在于如何判断no的情况，我只想到去一个比较大的数字来判断循环。

#include<iostream>
using namespace std;
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#define maxn 205
#define inf 0x3f3f3f3f
bool flag;
bool judge(bool x[],bool y[],int n,int m){
   for(int i=0;i<n;i++){
    if(x[i]==0)return false;
   }
   for(int i=0;i<m;i++){
    if(y[i]==0)return false;
   }
   flag=1;
   return true;
}
int main()
{
    int n,m;
    int b,g;
    int w;
    bool x[maxn],y[maxn];
    freopen("in.txt","r",stdin);
    while(cin>>n>>m){
        flag=0;
        memset(x,0,sizeof x);
        memset(y,0,sizeof y);
        cin>>b;
        for(int i=0;i<b;i++){
            cin>>w;
            x[w]=1;
        }
        cin>>g;
        for(int i=0;i<g;i++){
            cin>>w;
            y[w]=1;
        }
        int num=0;
        while(num<=n*m*1000){
            if(x[num%n]==1||y[num%m]==1){
                x[num%n]=y[num%m]=1;
            }
            if(judge(x,y,n,m))break;
            num++;
    /*
            if(num>=n*m*100000)break;
             for(int i=0;i<n;i++){
                 cout<<x[i]<<" ";
             }
              cout<<endl;
             for(int i=0;i<m;i++){
                cout<<y[i]<<" ";
             }
             cout<<endl;
             */
        }
        if(flag)printf("Yes\n");
        else printf("No\n");
    }
}

Drazil is playing a math game with Varda.

Let's define $\text{[math]}$ for positive integer x as a product of factorials of its digits. For example, $\text{[math]}$ .

First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:

1. x doesn't contain neither digit 0 nor digit 1.

2. $\text{[math]}$ = $\text{[math]}$ .

Help friends find such number.

Input

The first line contains an integer n (1 ≤ n ≤ 15) — the number of digits in a.

The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes.

Output

Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.

Sample test(s)

input
4
1234

output
33222

input
3
555

output
555

Note

In the first case, $\text{[math]}$

题意：找到和a!*b!*c!....等价的c!*d!*f!....的最大的cdf..

而且不能含有1或0。

思路：

因为只有1~9，所以分别看看1到9的阶乘能怎么拆，9的阶乘有点难猜，9！=9*8*7！，8*9=72=36*2=6*6*2=3！*3！*2！

用一个优先队列维护大的数在前面。。也可以最后从大到小排序输出。

#include<iostream>
using namespace std;
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#define maxn 205
#define inf 0x3f3f3f3f
typedef long long ll;
int main()
{
    int n;
    ll a;
    int c[16];
    int t;
    priority_queue<int>q;
    //freopen("in.txt","r",stdin);
    while(cin>>n){
        cin>>a;
        int i=1;
        t=0;
        while(i<=n){
            c[i]=a%10;
            i++;
            a=a/10;
        }
        for(int i=1;i<=n;i++){
            if(c[i]==1||c[i]==0)continue;
            else if(c[i]==4){
                q.push(3);
                q.push(2);
                q.push(2);
            }
            else if(c[i]==6){
                q.push(3);
                q.push(5);
            }
            else if(c[i]==8){
                q.push(2);
                q.push(2);
                q.push(2);
                q.push(7);
            }
            else if(c[i]==9){
                q.push(7);
                q.push(3);
                q.push(3);
                q.push(2);
            }
            else q.push(c[i]);
        }
        while(!q.empty()){
            cout<<q.top();
            q.pop();
        }
        cout<<endl;
    }
}