題意：要輸出N個數字a[N]，輸出的時候可以連續連續的輸出，每連續輸出一串，它的費用是 “這串數字和的平方加上一個常數M”。

析：這個題很容易想到DP方程dp[i] = min{dp[j] + M + (sum[i]-sum[j])^2}，但是很明顯是O(n^2)，TLE是必然的，所以要進行優化。

假設 i > j > k ，並且 j 要比 k 好，那麽就是 dp[j] + M + (sum[i]-sum[j])^2 < dp[k] + M + (sum[i]-sum[k])^2，

化簡 (dp[j]+sum[j]^2-(dp[k]+sum[k]^2))/(2*(sum[j]-sum[k]))<sum[i]，仔細看這就是一個斜率，這就說明了 j 比 k 要好，反之 j 比 k 要差。

然後就可以用單調隊列進行優化，要刪除一些不可能成為最優解的點。分別要隊首和隊尾進行優化。

代碼如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int dp[maxn], sum[maxn];
int q[maxn];

int DP(int i, int j){
  return dp[j] + m + (sum[i] - sum[j]) * (sum[i] - sum[j]);
}

int UP(int i, int j){
  return dp[i] + sum[i] * sum[i] - dp[j] - sum[j] * sum[j];
}

int DOWN(int i, int j){
  return 2 * (sum[i] - sum[j]);
}

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    dp[0] = 0;
    for(int i = 1; i <= n; ++i){
      int x;
      scanf("%d", &x);
      sum[i] = x + sum[i-1];
    }
    int fro = 0, rear = 0;
    q[++rear] = 0;
    for(int i = 1; i <= n; ++i){
      while(fro + 1 < rear && UP(q[fro+2], q[fro+1]) <= sum[i] * DOWN(q[fro+2], q[fro+1]))  ++fro;
      dp[i] = DP(i, q[fro+1]);
      while(fro + 1 < rear && UP(i, q[rear]) * DOWN(q[rear], q[rear-1]) <= UP(q[rear], q[rear-1]) * DOWN(i, q[rear]))  --rear;
      q[++rear] = i;
    }
    printf("%d\n", dp[n]);
  }
  return 0;
}

HDU 3507 Print Article (斜率DP)