【题解】SDOI2016征途

　　就放个代码吧……实在是太套路了。不过据说有复杂度还要低很多的算法，不知道是怎么做呀……

#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[][maxn], q[maxn], X[][maxn], Y[][maxn];

int read()
{
    int x = , k = ;
    char c;
    c = getchar();
    while(c < '' || c > '') { if(c == '-') k = -; c = getchar(); }
    while(c >= '' && c <= '') x = x *  + c - '', 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 = ; i <= n; i ++) sum[i] = read() + sum[i - ];
    int now = , pre = ;
    for(int i = ; i <= m; i ++)
    {
        memset(dp[now], -, sizeof(dp[now]));
        if(i != ) head = , tail = ;
        for(int j = ; j <= n; j ++)
        {
            while(head +  <= tail && Get_S(q[head], q[head + ], 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] =  * sum[j], Y[now][j] = dp[now][j] + sum[j] * sum[j];
            if(dp[pre][j] == -) continue;
            while(tail -  >= head && Get_S(q[tail], q[tail - ], 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 ;
}