C - Sonya and Problem Wihtout a Legend

思路：感觉没有做过这种套路题完全不会啊。。 把严格单调递增转换成非严格单调递增，所有可能出现的数字就变成了原数组出现过的数字。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std;

const int N = 3000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

LL dp[N][N], a[N], hs[N], tot, n;

int main() {
    memset(dp, INF, sizeof(dp));
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
        a[i] = a[i] - i;
        hs[tot++] = a[i];
    }
    sort(hs, hs + tot);
    tot = unique(hs, hs + tot) - hs;

    for(int j = 0; j < tot; j++) dp[1][j] = abs(a[1] - hs[j]);

    for(int i = 2; i <= n; i++) {
        LL mn = INF;
        for(int j = 0; j < tot; j++) {
            mn = min(mn, dp[i - 1][j]);
            dp[i][j] = min(dp[i][j], mn + abs(a[i] - hs[j]));
        }
    }
    LL ans = INF;
    for(int j = 0; j < tot; j++)
        ans = min(ans, dp[n][j]);
    printf("%lld\n", ans);
    return 0;
}

/*
*/