玲珑杯 1014 - Absolute Defeat-优快云博客

本文介绍了一个关于数组操作的算法问题，目标是最小化操作次数使数组中所有子序列的最小值不低于给定阈值。通过从左到右扫描数组并调整不足的子序列来解决该问题。

1014 - Absolute Defeat

Time Limit：2s Memory Limit：64MByte

Submissions：257Solved：73

DESCRIPTION

Eric has an array of integers 
a1,a2,...,ana1,a2,...,an.
 Every time, he can choose a contiguous subsequence of length 
kk
 and increase every integer in the contiguous subsequence by 
11.He
 wants the minimum value of the array is at least 
mm.
 Help him find the minimum number of operations needed.

INPUT

There are multiple test cases. The first line of input contains an integerTT,
 indicating the number of test cases. For each test case:The first line contains three integersnn,mm
 and kk(1≤n≤105,1≤k≤n,1≤m≤104)(1≤n≤105,1≤k≤n,1≤m≤104).The
 second line contains nn
 integers a1,a2,...,ana1,a2,...,an(1≤ai≤104)(1≤ai≤104).

OUTPUT

For each test case, output an integer denoting the minimum number of operations needed.

SAMPLE INPUT

32 2 21 15 1 41 2 3 4 54 10 31 2 3 4

SAMPLE OUTPUT

1015

【解题方法】思维。从左往右枚举每个元素, 如果不够的话, 直接把这个元素起始的长度为 $m$ 的序列都改一下就好了.

【AC 代码】

//
//Created by just_sort 2016/9/25 21:03
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
int a[1000010];
int main()
{
    int T,n,k,m;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1; i<=n; i++) scanf("%d",&a[i]);
        int ans = 0;
        int temp;
        for(int i=1; i<=n; i++){
            if(a[i]<m){
                temp = m - a[i];
                ans += temp;
                for(int j=i; j<i+k; j++){
                    a[j] += temp;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}