【题解】HAOI2008木棍分割

　　对于这道题目的两问，第一问直接二分答案求出最短长度。关键在于第二问应当如何求：建立dp方程，dp[i][j]代表到第i个分界线，切了j次（强制在第i处切一刀、这样就不会对后面的状态产生影响）。状态转移的方程即是当前分界线枚举上一条分界线在哪里，上一条分界线与当前线之间如果相差不超过之前二分出来的答案，就可以判定合法，方案数累加。因为注意到合法的分界线必然是一段连续区间，且单调右移不减，所以使用一个队列来维护队列内的元素总值。虽然让我感到非常玄学的是明明数据很小，我的数组一开小了就WA？不是很理解……

#include <bits/stdc++.h>
using namespace std;
#define maxn 5000000
#define maxm 300000
#define mod 10007
#define INF 99999999
int m, maxx, fans, ans, a[maxn], sum[maxn], dp[maxm][], n, pre = , now = ;
int q[maxn], tot, head, tail;
bool mark[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;
}

bool check(int x)
{
    int cnt = m;
    int tem = ;
    for(int i = ; i <= n; i ++)
    {
        if(tem + a[i] <= x) tem += a[i];
        else
        {
            if(!cnt) return false;
            tem = a[i], cnt --;
        }
    }
    return true;
}

void solve()
{
    int l = maxx, r = sum[n];
    while(l <= r)
    {
        int mid = (l + r) >> ;
        if(check(mid)) ans = mid, r = mid - ;
        else l = mid + ;
    }
}

void DP()
{
    dp[][pre] = ;
    for(int i = n - ; i >= ; i --)
    {
        if(sum[n] - sum[i] <= ans)
            mark[i] = true;
        else break;
    }
    bool flag = false;
    for(int i = ; i <= m; i ++)
    {
        head = , tail = ;
        q[++ tail] = i - , tot = dp[i - ][pre];
        for(int j = i; j <= n; j ++)
        {
            while(head <= tail && sum[j] - sum[q[head]] > ans)
                tot = (tot - dp[q[head]][pre] + mod) % mod, head ++;
            dp[j][now] = tot, dp[j][now] %= mod;
            if(dp[j][pre]) q[++ tail] = j, tot += dp[j][pre], tot %= mod;
            if(mark[j])
            {
                fans += dp[j][now];
                fans %= mod;
            }
        }
        now ^= , pre ^= ;
    }
}

int main()
{
    n = read(), m = read();
    for(int i = ; i <= n; i ++)
    {
        a[i] = read(), sum[i] = sum[i - ] + a[i];
        maxx = max(a[i], maxx);
    }
    solve();
    cout << ans << " ";
    DP();
    cout <<fans << endl;
    return ;
}