Description

题库链接

有 \(n\) 根木棍，第 \(i\) 根木棍的长度为 \(L_i\) ， \(n\) 根木棍依次连结了一起，总共有 \(n-1\) 个连接处。现在允许你最多砍断 \(m\) 个连接处，砍完后 \(n\) 根木棍被分成了很多段，要求满足总长度最大的一段长度最小，并且输出有多少种砍的方法使得总长度最大的一段长度最小。对质数取模。

\(1\leq n\leq 50000,1\leq m\leq \min\{n-1,1000\}\)

Solution

第一问二分，不再赘述。

第二问考虑 \(DP\) 。令 \(f_{i,j}\) 为前 \(i\) 条木棍，划分为 \(j\) 段的方案数。转移我们考虑最后一段怎么分。我们可以预处理出一个数组 \(pre_i\) 表示最后一段最长能分到哪里，显然 \[f_{i,j}=\sum_{k=pre_i}^{i-1}f_{k,j-1}\]

显然这样复杂度是假的。我们可以用前缀和优化，并且珂以记 \(f_{i}\) 为当前枚举到分为 \(j\) 段时，第 \(i\) 条木棍前的方案数。这样直接省去一维。

空间复杂度 \(O(n)\) ；时间复杂度 \(O(nm)\) 。

Code

//It is made by Awson on 2018.3.3

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 50000, INF = 5e8, yzh = 10007;

void read(int &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, m, a[N+5], len, lef[N+5], f[N+5], s[N+5];

bool judge(int x) {

    int cnt = 1, ret = 0;

    for (int i = 1; i <= n; i++) {

    if (a[i] > x) return false;

    if (ret+a[i] > x) ret = a[i], ++cnt; else ret += a[i];

    }

    return cnt <= m;

}

void solve1() {

    int l = 1, r = INF, ans;

    while (l <= r) {

    int mid = (l+r)>>1; if (judge(mid)) ans = mid, r = mid-1; else l = mid+1;

    }

    write(len = ans), putchar(' ');

}

void solve2() {

    int now = 0, ans = 0;

    for (int i = 1; i <= n; i++) {

    a[i] += a[i-1];

    while (a[i]-a[now] > len) ++now;

    lef[i] = now;

    }

    for (int i = 0; i <= n; i++) s[i] = 1;

    for (int len = 1; len <= m; len++) {

    for (int i = 1; i <= n; i++) f[i] = (lef[i] == 0 ? s[i-1] : s[i-1]-s[lef[i]-1]);

    s[0] = 0; for (int i = 1; i <= n; i++) s[i] = (f[i]+s[i-1])%yzh;

    ans = (ans+f[n])%yzh;

    }

    write((ans+yzh)%yzh);

}

void work() {

    read(n), read(m); ++m; for (int i = 1; i <= n; i++) read(a[i]);

    solve1(); solve2();

}

int main() {

    work(); return 0;

}