FacePalm Accounting

水题：给定n个数，要求任意k个数的和都为负数，对于每个数的操作不能减到小于初始的最小值，更改的代价为减小的值的和，问最小代价。
贪心
从左到右对于每段序列，考虑从右往左进行改变，因为这样对于后面的增益是最大的，然后减到最小值再往前找不为最小值的数再进行减操作。我用了链表优化

#include <stdio.h>
#include <stdlib.h>

const int maxn = 300000;
int a[maxn], pre[maxn], n, m;

int main() {
    scanf("%d%d", &n, &m);
    int minn = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (a[i] < minn) minn = a[i];
        pre[i] = i-1;
    }
    int sum = 0, ans = 0;
    a[0] = 0;
    for (int i = 1; i < m; i++) sum += a[i];
    for (int i = m; i <= n; i++) {
        sum = sum+a[i]-a[i-m];
        if (sum >= 0) {
            for (int pos = i; pos >= 1; pos = pre[pos]) {
                if (sum-(a[pos]-minn) < 0) {
                    int tar = a[pos]-1-sum;
                    ans += a[pos]-tar;
                    a[pos] = tar;
                    sum = -1;
                    pre[i+1] = pos;
                    break;
                } else {
                    ans += a[pos]-minn;
                    sum = sum-a[pos]+minn;
                    a[pos] = minn;
                }
            }
        }
    }
    printf("%d\n", ans);
    for (int i = 1; i < n; i++) printf("%d ", a[i]);
    printf("%d\n", a[n]);
}