【水贪心】#11 A. Increasing Sequence

最新推荐文章于 2021-03-19 03:37:59 发布

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本文探讨了一种用于将给定序列转换为递增序列的算法，通过使用除法优化，实现序列调整所需的最少步骤。该算法适用于具有特定限制条件的场景，包括序列长度和增量值。详细介绍了算法原理、实现细节以及实例分析，旨在提高处理此类问题的效率。

A sequence a0, a1, ..., at - 1 is called increasing if ai - 1 < ai for each i: 0 < i < t.

You are given a sequence b0, b1, ..., bn - 1 and a positive integer d. In each move you may choose one element of the given sequence and add d to it. What is the least number of moves required to make the given sequence increasing?

Input

The first line of the input contains two integer numbers n and d (2 ≤ n ≤ 2000, 1 ≤ d ≤ 106). The second line contains space separated sequence b0, b1, ..., bn - 1 (1 ≤ bi ≤ 106).

Output

Output the minimal number of moves needed to make the sequence increasing.

Sample test(s)

input
4 2
1 3 3 2

output
3

一道模拟题，其实直接模拟就出了，不过用一下除法来提升效率还是养成习惯的好。

贪心：遍历一次，每个当前数比自己前一个数大即可

C++：

#include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
	int n,d,cnt=0;
	scanf("%d%d",&n,&d);
	int val[n];
	for(int i=0;i<n;i++)
		scanf("%d",&val[i]);
	
	for(int i=1;i<n;i++)
	{
			if(val[i]>val[i-1])continue;
			else 
			{
				int temp=val[i-1]-val[i];
				val[i]+=(temp/d+1)*d;
				cnt+=(temp/d+1);
			}
		
	}
	printf("%d",cnt);
	return 0;
}