做题记录 - CF1175D Array Splitting

洛谷题面

题面

将 $n$ 个正整数不调整顺序分成 $k$ 段，编号$1\dots k$ ，使得$\sum _{i=1}^n{a}_i\cdot f\left(i\right)$最大。
我们规定 $f\left(i\right)$ 表示第 $i$ 个数属于的段编号。

心路历程

Round 1

$$\begin{array}{r}\sum _{i=1}^n{a}_i\cdot f\left(i\right)\to a_1+a_2\cdot f\left(2\right)+a_3\cdot f\left(3\right)+\cdots +a_n\cdot f\left(n\right)\end{array}$$

由于相同的段中元素的 $f\left(\right)$ 相同，提取公因数，得：

$$a_1+a_2+\cdots +a_i+(a_{i+1}+a_{i+2}+\cdots +a_j)\times 2+\cdots +(a_k+a_{k+1}+\cdots +a_n)\times k$$

拆括号，得：

$$a_1+a_2+\cdots +a_i+2a_{i+1}+2a_{i+2}+\cdots +2a_j+\cdots +k{a}_k+k{a}_{k+1}+\cdots +k{a}_n$$

思路中断。

Round 2

考虑分段求和，$su{m}_i$ 为第 $i$ 段的和（考虑前缀和，后文 $s_i$ 即前缀和数组），即：

$$ans=\sum _{i=1}^{k}i\times su{m}_i$$

把$\sum$ 展开，得：

$$ans=s_1+2(s_2-s_1)+\cdots +(k-1)(s_{k-1}-s_{k-2})+k(s_k-s_{k-1})$$

继续化简，得：

$$\begin{array}{r}ans=s_1-2s_1+2s_2-3s_2+\cdots +(k-1)s_{k-1}-k{s}_{k-1}+k{s}_k\\ =-(s_1+s_2+\cdots +s_{k-1})+k{s}_k\\ =k{s}_k-\sum _{i=1}^{k-1}{s}_i\end{array}$$

时间复杂度 $O(2n)$
空间复杂度 $O(2n)$

跑样例后发现WA了

核心代码：

    read(n, k);
    for(int i = 1; i <= n; ++i) {
        read(a[i]);
        s[i] = s[i - 1] + a[i];
    }
    for(int i = 1; i < k; ++i) tot += s[i];
    cout << k * s[k] - tot;

Round DEBUG

检查推导，发现推导的第二步不合理。

Round 3

再推导一遍。

$$\begin{array}{r}\sum _{i=1}^n{a}_i\cdot f\left(i\right)\\ =a_1+\cdots +a_{x_1}+2(a_{x_1}+a_{x_1+1}+\cdots +a_{x_2})\cdot 2+\cdots +(a_{x_{k-1}}+\cdots +a_n)\cdot k\\ =s_{x_1}+2(s_{x_2}-s_{x_1})+\cdots +(k-1)(s_{x_{k-1}}-s_{x_{k-2}})+k(s_n-s_{x_{k-1}})\\ =s_{x_1}+2s_{x_2}-2s_{x_1}+\cdots +(k-1)s_{x_3}-(k-1)s_{x_{k-2}}+k{s}_n-k{s}_{x_{k-1}}\\ =k{s}_n-s_{x_1}-s_{x_2}-\cdots -s_{x_{k-1}}\end{array}$$

想要让这个式子最大，只需要让$s_{x_1}+s_{x_2}+\cdots +s_{x_{k-1}}$ 最小即可所以对前缀和数组排序，取第$1\to k-1$ 个即可。注意排序后不影响题面所说

不能动原始序列的顺序

$$\begin{array}{r}\therefore \mathrm{排}\mathrm{序}\mathrm{后}\mathrm{求}k{s}_n+\sum _{i=1}^{k-1}{s}_i\mathrm{即}\mathrm{可}\end{array}$$

核心代码：

	read(n, k);
    for(int i = 1; i <= n; ++i) read(a[i]);
    for(int i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i];
    sort(s + 1, s + n);
    for(int i = 1; i < k; ++i) tot += s[i];
    write(k * s[n] - tot);

WA

Round DEBUG

算了一下数据范围，会爆int

十年OI一场空，不开long long见祖宗

AC Code

#pragma region
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <cmath>
#include <cstring>
#include <type_traits>

using namespace std;
using ll = long long;
using LL = ll;
const int INF = 0x7fffffff;
inline void IOS() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
}

template <typename T>
void read(T &x) {
	x = 0; char c = getchar(); int f = 0;
	for (; !isdigit(c); c = getchar())
		f |= c == '-';
	for (; isdigit(c); c = getchar())
		x = x * 10 + (c ^ '0');
	if (f) x = -x;
}
template <typename T, typename ... Args>
void read(T& a, Args&... args) {
    read(a), read(args...);
}

template <typename T>
void write(T x, char ed = '\n') {
    if(std::is_same<typename std::decay<T>, char>::value) {
        putchar(x); return ;
    }
	if (x < 0) putchar('-'), x = -x;
	static short st[30], tp;
	do st[++tp] = x % 10, x /= 10; while (x);
	while (tp) putchar(st[tp--] | '0');
	putchar(ed);
}

template <typename T, typename ... Args>
void write(T& a, Args&... args) {
    write(a), write(args...);
}

#pragma endregion

const int N = 5e5 + 10;

void Solve();

ll n, k;
vector<ll> a(N);
ll s[N];
ll tot = 0;

signed main() {
    IOS();
    Solve();
    return 0;
}

void Solve_throw() {
    read(n, k);
    for(ll i = 1; i <= n; ++i) {
        read(a[i]);
        s[i] = s[i - 1] + a[i];
    }
    sort(s + 1, s + n);
    for(ll i = 1; i < k; ++i) tot += s[i];
    cout << k * s[n] - tot;
}

void Solve() {
    read(n, k);
    for(int i = 1; i <= n; ++i) read(a[i]);
    for(int i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i];
    sort(s + 1, s + n);
    for(int i = 1; i < k; ++i) tot += s[i];
    write(k * s[n] - tot);
}

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