[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$\left \lfloor \frac{k}{c}\right \rfloor$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 + ;
 ll dp[maxn], lg[maxn], Min[maxn][], a[maxn], sum[maxn];
 ll Q(int l, int r) {
     int k = lg[r - l + ];
     return min(Min[l][k], Min[r - ( << k) + ][k]);
 }
 int main() {
     int n, m;
     scanf("%d%d", &n, &m);
     for (int i = ; i <= n; i++)
         scanf("%lld", &a[i]), sum[i] = sum[i - ] + a[i];
     lg[] = -;
     for (int i = ; i <= n; i++) {
         if ((i & (i - )) == )
             lg[i] = lg[i - ] + ;
         else
             lg[i] = lg[i - ];
     }
     for (int i = ; i <= n; i++)
         Min[i][] = a[i];
     for (int j = ; ( << j) <= n; j++)
         for (int i = ; i + ( << j) -  <= n; i++)
             Min[i][j] = min(Min[i][j - ], Min[i + ( << (j - ))][j - ]);
     memset(dp, , sizeof(dp));
     dp[] = ;
     for (int i = ; i <= n; i++) {
         dp[i] = dp[i - ]+a[i];
         if (i - m >= )
             dp[i] = min(dp[i], dp[i - m] + sum[i] - sum[i - m] - Q(i - m + , i));
     }
     printf("%lld\n", dp[n]);
 }