Codeforces Beta Round #11 A. Increasing Sequence 贪心

A. Increasing Sequence

题目连接:

http://www.codeforces.com/contest/11/problem/A

Description

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 Input

4 2
1 3 3 2

Sample Output

3

Hint

题意

你每次操作可以使得一个数增加d,问你最小操作多少次,可以使得这个序列变成一个递增序列

题解:

贪心就好了,能变就变,不用变的时候不变就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
int a[maxn];

int main()
{
    int n,d;
    scanf("%d%d",&n,&d);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    long long ans = 0;
    for(int i=2;i<=n;i++)if(a[i]<=a[i-1])
    {
        int tmp = (a[i-1]-a[i]+d)/d;
        a[i]+=tmp*d;
        ans+=tmp;
    }
    cout<<ans<<endl;
}
    原文作者:qscqesze
    原文地址: https://www.cnblogs.com/qscqesze/p/5440181.html
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