CodeForces - 11A ---递增数列

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我是靠谱客的博主秀丽期待，最近开发中收集的这篇文章主要介绍CodeForces - 11A ---递增数列，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

#include<cstdio>
#include<cstring>
#include<string>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<iostream>
#define ll long long
using namespace std;

int a[2010];
int n, d;
int main(){
    scanf("%d%d", &n, &d);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int ans = 0;
    for(int i = 1; i <= n; i++){
        if(a[i] <= a[i-1]){
            int temp = (a[i-1] - a[i] + d) / d;
            a[i] = a[i] + temp*d;
            ans += temp;
        }
    }
    cout << ans << endl;
    return 0;
}

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

You are given a sequence b₀, b₁, ..., b_n - 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, ..., b_n - 1 (1 ≤ b_i ≤ 10⁶).