[codeforces940E]Cashback

题目链接

题意是说将$n$个数字分段使得每段贡献之和最小，每段的贡献为区间和减去前$left lfloor frac{k}{c} ight floor$小的和。

仔细分析一下可以知道，减去$2$个可以分成减去$2$次$1$个，所以就可以设一个$dp:$$dp[i]$为$1-i$位的最小和.

$dp[i]=dp[i-1]+a[i]$，表示第$i$个单独分成一组。

$dp[i]=dp[i-m]+sum[i]-sum[i-m]-Q(i-m+1,i)$，表示第$i-c$到第$i$个分成一组，就要减去区间内的最小值。

所以ST表预处理一下

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
ll dp[maxn], lg[maxn], Min[maxn][20], a[maxn], sum[maxn];
ll Q(int l, int r) {
    int k = lg[r - l + 1];
    return min(Min[l][k], Min[r - (1 << k) + 1][k]);
}
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%lld", &a[i]), sum[i] = sum[i - 1] + a[i];
    lg[0] = -1;
    for (int i = 1; i <= n; i++) {
        if ((i & (i - 1)) == 0)
            lg[i] = lg[i - 1] + 1;
        else
            lg[i] = lg[i - 1];
    }
    for (int i = 1; i <= n; i++)
        Min[i][0] = a[i];
    for (int j = 1; (1 << j) <= n; j++)
        for (int i = 1; i + (1 << j) - 1 <= n; i++)
            Min[i][j] = min(Min[i][j - 1], Min[i + (1 << (j - 1))][j - 1]);
    memset(dp, 127, sizeof(dp));
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
        dp[i] = dp[i - 1]+a[i];
        if (i - m >= 0)
            dp[i] = min(dp[i], dp[i - m] + sum[i] - sum[i - m] - Q(i - m + 1, i));
    }
    printf("%lld
", dp[n]);
}