codeforces 360 B

B - Levko and Array

题目大意：给你你个长度为n的数列a，你最多改变k个值，max{ abs ( a[ i + 1] - a[ i ] ) } 的最小值为多少。

思路：这个题很难想到如何取check。。 二分最小值，然后用dp进行check，dp[ i ]表示前 i 项中第 i 个不改变最少

需要改变几个值。 

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 2000 + 7;
const int M = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int n, k, dp[N];
int a[N];

bool check(LL mx) {
    for(int i = 1; i <= n; i++) dp[i] = i - 1;
    for(int i = 2; i <= n; i++) {
        for(int j = 1; j < i; j++) {
            if(abs(a[i] - a[j]) <= mx * (i - j)) {
                dp[i] = min(dp[i], dp[j] + i - j - 1);
            }
        }
    }
    for(int i = 1; i <= n; i++)
        if(dp[i] + n - i <= k) return true;
    return false;
}

int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    LL l = 0, r = 2e9, mid, ans = 2e9;
    while(l <= r) {
        mid = l + r >> 1;
        if(check(mid)) r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}


/*
*/