就放個代碼吧……實在是太套路了。不過據說有復雜度還要低很多的算法，不知道是怎麽做呀……

#include <bits/stdc++.h>
using namespace std;
#define maxn 10000
#define int long long
#define db double
int n, m, head, tail, sum[maxn];
int dp[2][maxn], q[maxn], X[2][maxn], Y[2][maxn];

int read()
{
    int x = 0, k = 1;
    char c;
    c = getchar();
    
 
while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) k = -1; c = getchar(); }
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * k;
}

db Get_S(int a, int b, int opt)
{
    return (db) (Y[opt][a] - Y[opt][b]) / (db) (X[opt][a] - X[opt][b]);
}

signed main()
{
    n 
 
= read(); m = read();
    for(int i = 1; i <= n; i ++) sum[i] = read() + sum[i - 1];
    int now = 1, pre = 0;
    for(int i = 1; i <= m; i ++)
    {
        memset(dp[now], -1, sizeof(dp[now]));
        if(i != 1) head = 1, tail = 0;
        for(int j = 1; j <= n; j ++)
        {        
            
 
while(head + 1 <= tail && Get_S(q[head], q[head + 1], pre) < (db) sum[j]) head ++;
            if(head <= tail) dp[now][j] = dp[pre][q[head]] + (sum[j] - sum[q[head]]) * (sum[j] - sum[q[head]]);
            X[now][j] = 2 * sum[j], Y[now][j] = dp[now][j] + sum[j] * sum[j];
            if(dp[pre][j] == -1) continue;
            while(tail - 1 >= head && Get_S(q[tail], q[tail - 1], pre) > Get_S(q[tail], j, pre)) tail --;
            q[++ tail] = j;
        }
        swap(now, pre);
    }
    int ans = dp[pre][n] * m - sum[n] * sum[n];
    printf("%lld\n", ans);
    return 0;
}

【題解】SDOI2016征途