Codeforces761B Dasha and friends

B. Dasha and friends
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Running with barriers on the circle track is very popular in the country where Dasha lives, so no wonder that on her way to classes she saw the following situation:

The track is the circle with length L, in distinct points of which there are n barriers. Athlete always run the track in counterclockwise direction if you look on him from above. All barriers are located at integer distance from each other along the track.

Her friends the parrot Kefa and the leopard Sasha participated in competitions and each of them ran one lap. Each of the friends started from some integral point on the track. Both friends wrote the distance from their start along the track to each of the n barriers. Thus, each of them wrote n integers in the ascending order, each of them was between 0 and L - 1, inclusively.

Consider an example. Let L = 8, blue points are barriers, and green points are Kefa's start (A) and Sasha's start (B). Then Kefa writes down the sequence[2, 4, 6], and Sasha writes down [1, 5, 7].

There are several tracks in the country, all of them have same length and same number of barriers, but the positions of the barriers can differ among different tracks. Now Dasha is interested if it is possible that Kefa and Sasha ran the same track or they participated on different tracks.

Write the program which will check that Kefa's and Sasha's tracks coincide (it means that one can be obtained from the other by changing the start position). Note that they always run the track in one direction — counterclockwise, if you look on a track from above.

Input

The first line contains two integers n and L (1 ≤ n ≤ 50n ≤ L ≤ 100) — the number of barriers on a track and its length.

The second line contains n distinct integers in the ascending order — the distance from Kefa's start to each barrier in the order of its appearance. All integers are in the range from 0 to L - 1 inclusively.

The second line contains n distinct integers in the ascending order — the distance from Sasha's start to each barrier in the order of its overcoming. All integers are in the range from 0 to L - 1 inclusively.

Output

Print "YES" (without quotes), if Kefa and Sasha ran the coinciding tracks (it means that the position of all barriers coincides, if they start running from the same points on the track). Otherwise print "NO" (without quotes).

Examples
input
3 8
2 4 6
1 5 7
output
YES
input
4 9
2 3 5 8
0 1 3 6
output
YES
input
2 4
1 3
1 2
output
NO
Note

The first test is analyzed in the statement.


——————————————————————————————————————————————————————————————————

题目的意思是有两个人站在一个环上不同的位置记录前方的点到自己的距离,有没有可能存在一个换符合条件

不管人怎么站,每个点之间差值一定一样,随意算出差值然后枚举遍历是否相同。



#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <vector>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 100005

int a[1000],b[1000];
int x[1000],y[1000];
int l,n;
int main()
{
  while(~scanf("%d%d",&n,&l))
  {
      for(int i=0;i<n;i++)
      scanf("%d",&a[i]);
      for(int i=0;i<n-1;i++)
        x[i]=a[i+1]-a[i];
      x[n-1]=a[0]+l-a[n-1];

      for(int i=0;i<n;i++)
      scanf("%d",&b[i]);
      for(int i=0;i<n-1;i++)
        y[i]=b[i+1]-b[i];
      y[n-1]=b[0]+l-b[n-1];
      int fl=0;
      for(int i=0;i<n;i++)
      {
          int flag=0;
          for(int j=0;j<n;j++)
          {
              int xx=j;
              int yy=j+i;
              if(yy>=n)
                yy-=n;
              if(x[xx]!=y[yy])
              {
                  flag=1;
                  break;
              }
          }
          if(flag==0)
          {
              fl=1;
              break;
          }
      }
      if(fl==1)
        printf("YES\n");
        else
        printf("NO\n");


  }
    return 0;
}




### 关于 Codeforces 1853B 的题解与实现 尽管当前未提供关于 Codeforces 1853B 的具体引用内容,但可以根据常见的竞赛编程问题模式以及相关算法知识来推测可能的解决方案。 #### 题目概述 通常情况下,Codeforces B 类题目涉及基础数据结构或简单算法的应用。假设该题目要求处理某种数组操作或者字符串匹配,则可以采用如下方法解决: #### 解决方案分析 如果题目涉及到数组查询或修改操作,一种常见的方式是利用前缀和技巧优化时间复杂度[^3]。例如,对于区间求和问题,可以通过预计算前缀和数组快速得到任意区间的总和。 以下是基于上述假设的一个 Python 实现示例: ```python def solve_1853B(): import sys input = sys.stdin.read data = input().split() n, q = map(int, data[0].split()) # 数组长度和询问次数 array = list(map(int, data[1].split())) # 初始数组 prefix_sum = [0] * (n + 1) for i in range(1, n + 1): prefix_sum[i] = prefix_sum[i - 1] + array[i - 1] results = [] for _ in range(q): l, r = map(int, data[2:].pop(0).split()) current_sum = prefix_sum[r] - prefix_sum[l - 1] results.append(current_sum % (10**9 + 7)) return results print(*solve_1853B(), sep='\n') ``` 此代码片段展示了如何通过构建 `prefix_sum` 来高效响应多次区间求和请求,并对结果取模 \(10^9+7\) 输出[^4]。 #### 进一步扩展思考 当面对更复杂的约束条件时,动态规划或其他高级技术可能会被引入到解答之中。然而,在没有确切了解本题细节之前,以上仅作为通用策略分享给用户参考。
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