cf 443(div2) 部分题目

http://codeforces.com/contest/879

It seems that Borya is seriously sick. He is going visit n doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3and so on). Borya will get the information about his health from the last doctor.

Doctors have a strange working schedule. The doctor i goes to work on the si-th day and works every di day. So, he works on days si, si + di, si + 2di, ....

The doctor's appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors?

Input

First line contains an integer n — number of doctors (1 ≤ n ≤ 1000).

Next n lines contain two numbers si and di (1 ≤ si, di ≤ 1000).

Output

Output a single integer — the minimum day at which Borya can visit the last doctor.

Examples

input

Copy

output

input

Copy

2
10 1
6 5

output

Note

In the first sample case, Borya can visit all doctors on days 2, 3 and 4.

In the second sample case, Borya can visit all doctors on days 10 and 11.

题意大概就是每个医生出诊时间是等差数列问把每一个医生都看遍最少需要多少天

思路也很简单随便找一个人加上公差看在不在下一个天数范围内在就不用更新不在就更新一下天数

#include<bits/stdc++.h>

using namespace std;
#define MAXN 1000

char mp[1005][1005];
int Next[4][2]= {0,1,1,0,0,-1,-1,0},book[1000005][2],vis[1005][1005],n,m=0,a,b,sum;


int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a>>b;
        if(m<a)
            m=a;
        else if(m>=a)
        {
            while(m>=a)
                a+=b;
            m=a;
        }
    }
    cout<<m<<endl;
    return 0;
}

ac代码

n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.

For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.

Input

The first line contains two integers: n and k (2 ≤ n ≤ 500, 2 ≤ k ≤ 1012) — the number of people and the number of wins.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.

Output

Output a single integer — power of the winner.

Examples

input

Copy

2 2
1 2

output

input

Copy

4 2
3 1 2 4

output

input

Copy

6 2
6 5 3 1 2 4

output

input

Copy

2 10000000000
2 1

output

Note

Games in the second sample:

3 plays with 1. 3 wins. 1 goes to the end of the line.

3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.

题意就是排队打球限定回合数找出在限定回合数内最强的人

思路这题最好别用队列做我第一反应想这是个裸队列随便写写就好了结果花式gg 不如就用数组记录k次下谁最大而且 n<k 最大的一定是力量最大的不然的话如果一旦有人输了那么他一定不会是最大的所以只需要维护一直在队列首的那个人的power就行了

#include<bits/stdc++.h>

using namespace std;
#define MAXN 1000

int Next[4][2]= {0,1,1,0,0,-1,-1,0},book[1000],n,x,y,s=0;
long long int k;
int a[1000];

int main()
{
    cin>>n>>k;
    memset(book,0,sizeof(book));
    for(int i=0; i<n; i++)
        cin>>a[i];
    int p=a[0];
    for(int i=1; i<n; i++)
    {
        if(s>=k)
            break;
        if(p>a[i])
            s++;
        else
        {
            s=1;
            p=a[i];
        }
    }
    cout<<p;
    return 0;
}

View Code