Codeforces Beta Round #11 A. Increasing Sequence 贪心
題目連接:
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
題意
你每次操作可以使得一個數(shù)增加d,問你最小操作多少次,可以使得這個序列變成一個遞增序列
題解:
貪心就好了,能變就變,不用變的時候不變就好了
代碼
#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; }總結(jié)
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