codeforceA. Piles With Stones【水题】

探讨在Innopolis的石头花园中,两组参与者记录的石头数量变化是否可能由EJOI评审团成员的夜间活动引起。评审团成员可能取走石头或将石头从一堆移到另一堆,参与者试图验证笔记的一致性。

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A. Piles With Stones

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a beautiful garden of stones in Innopolis.

Its most beautiful place is the nn piles with stones numbered from 11 to nn.

EJOI participants have visited this place twice.

When they first visited it, the number of stones in piles was x1,x2,…,xnx1,x2,…,xn, correspondingly. One of the participants wrote down this sequence in a notebook.

They visited it again the following day, and the number of stones in piles was equal to y1,y2,…,yny1,y2,…,yn. One of the participants also wrote it down in a notebook.

It is well known that every member of the EJOI jury during the night either sits in the room 108108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.

Participants want to know whether their notes can be correct or they are sure to have made a mistake.

Input

The first line of the input file contains a single integer nn, the number of piles with stones in the garden (1≤n≤501≤n≤50).

The second line contains nn integers separated by spaces x1,x2,…,xnx1,x2,…,xn, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0≤xi≤10000≤xi≤1000).

The third line contains nn integers separated by spaces y1,y2,…,yny1,y2,…,yn, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0≤yi≤10000≤yi≤1000).

Output

If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).

Examples

input

Copy

5
1 2 3 4 5
2 1 4 3 5

output

Copy

Yes

input

Copy

5
1 1 1 1 1
1 0 1 0 1

output

Copy

Yes

input

Copy

3
2 3 9
1 7 9

output

Copy

No

水题直接上代码,不要多想这个题

代码:

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
int a[108],b[108],a1,b1;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
          cin>>a[i];
          a1+=a[i];
    }

    for(int i=1;i<=n;i++)
    {
            cin>>b[i];
            b1+=b[i];
    }
    if(b1-a1<=2&&b1<=a1) printf("Yes\n");
    else printf("No\n");
    return 0;

}

 

### 关于算法设计的相关目 以下是关于算法设计的一些经典目及其解答思路: #### 目一:贪心算法中的最小费用合并问类似于哈夫曼编码的思想,目标是最小化合并成本。通过将输入数据按升序排列并始终选取前 $ k $ 堆进行合并可以实现最优解[^1]。 ```python import heapq def min_cost_to_merge_piles(piles, k): if not piles or len(piles) < 2: return 0 heap = [] for pile in piles: heapq.heappush(heap, pile) total_cost = 0 while len(heap) >= k: current_sum = sum([heapq.heappop(heap) for _ in range(k)]) total_cost += current_sum heapq.heappush(heap, current_sum) return total_cost ``` #### 目二:分治法求解最大子数组和 该问是经典的动态规划与分治方法结合的应用案例。其核心思想是利用递归来分割原数组,并计算跨越中间位置的最大子数组和[^2]。 ```java class Solution { public int maxSubArray(int[] nums) { return findMaxSubArray(nums, 0, nums.length - 1); } private int findMaxSubArray(int[] nums, int left, int right){ if (left == right) return nums[left]; int mid = (right - left) / 2 + left; int maxLeftSum = findMaxSubArray(nums, left, mid); int maxRightSum = findMaxSubArray(nums, mid + 1, right); // 计算跨中点的最大子数组和 int maxCrossingSum = calculateMaxCrossingSum(nums, left, mid, right); return Math.max(Math.max(maxLeftSum, maxRightSum), maxCrossingSum); } private int calculateMaxCrossingSum(int[] nums, int left, int mid, int right){ int leftSum = Integer.MIN_VALUE; int sum = 0; for (int i = mid; i >= left; --i){ sum += nums[i]; if (sum > leftSum){ leftSum = sum; } } int rightSum = Integer.MIN_VALUE; sum = 0; for (int j = mid + 1; j <= right; ++j){ sum += nums[j]; if (sum > rightSum){ rightSum = sum; } } return leftSum + rightSum; } } ``` 以上两种类型的目分别代表了 **贪心算法** 和 **分治策略** 的典型应用实例。 ---
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