P5686 [CSP-S2019 江西] 和积和 题解

PART 1： 题目大意

设 \(s(l,r) = \sum_{i = l}^{r} a_i \times b_i\)，求 \(\sum_{l=1}^{n}\sum_{r=1}^{n} s(l,r)\)

PART 2：解题思路

从数据范围可以看出来，这是一道思维题，我没需要把体面给出的柿子通过转化使其复杂度下降。从数据范围看出，这里需要时间复杂度大约 \(O(n)\)。

不妨先把题目给的柿子拆开来看是否存在规律：

\[ans = a_1 \times (\sum_{i = 1}^{n} b_i + \sum_{i = 1}^{n-1} b_i + \sum_{i = 1}^{n-2} b_i …… \sum_{i = 1}^{1} b_i) \\+ a_2 \times (\sum_{i = 1}^{n} b_i + \sum_{i = 1}^{n-1} b_i + \sum_{i = 1}^{n-2} b_i …… \sum_{i = 1}^{2} b_i+\sum_{i = 2}^{n} b_i + \sum_{i = 2}^{n-1} b_i + \sum_{i = 2}^{n-2} b_i …… \sum_{i = 2}^{2} b_i) \\ …… \]

设 \(a_m\) 对应的括号中的柿子为 \(f(m)\)，则：

\[f(m) = \sum_{i = 1}^{n} b_i + \sum_{i = 1}^{n-1} b_i + \sum_{i = 1}^{n-2} b_i …… \sum_{i = 1}^{m} b_i\\+\sum_{i = 2}^{n} b_i + \sum_{i = 2}^{n-1} b_i + \sum_{i = 2}^{n-2} b_i …… \sum_{i = 2}^{m} b_i \\ ……\\+\sum_{i = m}^{n} b_i + \sum_{i = m}^{n-1} b_i + \sum_{i = m}^{n-2} b_i …… \sum_{i = m}^{m} b_i \]

这样看上去貌似没有什么规律，我们举两个例子来看

\[f(1) = \sum_{i = 1}^{n} b_i + \sum_{i = 1}^{n-1} b_i + \sum_{i = 1}^{n-2} b_i …… \sum_{i = 1}^{1} b_i = \sum_{i = 1}^n b_i \times (n-i+1) \]

\[f(2) = \sum_{i = 1}^{n} b_i + \sum_{i = 1}^{n-1} b_i + \sum_{i = 1}^{n-2} b_i …… \sum_{i = 1}^{2} b_i+\sum_{i = 2}^{n} b_i + \sum_{i = 2}^{n-1} b_i + \sum_{i = 2}^{n-2} b_i …… \sum_{i = 2}^{2} b_i \\= \sum_{i = 1}^n b_i \times (n-i+1) - b_1 + \sum_{i = 2}^n b_i \times (n-i+1) \]

\[…… \]

感觉他们都有很相似的 \(\sum_{i = k}^n b_i \times (n-i+1)\) 这部分，不妨设 \(sum = \sum_{i = 1}^n b_i \times (n-i+1)\)，再用 \(sum\) 将其替换，那么会得到：

\[f(1) = sum \]

\[f(2) = 2sum - (n+1) b_1 \]

\[f(3) = 3sum - 2(n+1)b_1-(n-1)b_2 \]

以此类推，那么得到：

\[f(m) = m\times sum - \sum_{i=1}^{m-1}(n+1)b_1\times (m-i) \]

这样的柿子是 \(O(n^2)\) 的，于是我们就拿到了 \(70pts\) 的分数，代码如下：

#include<bits/stdc++.h>
#define MAXN 500010
#define MOD 1000000007ll
using namespace std;
typedef long long ll;
int n;
ll a[MAXN],b[MAXN],sum,ans;
int main(){
    // freopen("sum.in","r",stdin);
    // freopen("sum.out","w",stdout);
    scanf("%d",&n);
    for(int i = 1;i <= n;i++) scanf("%lld",&a[i]), a[i] %= MOD;
    for(int i = 1;i <= n;i++) scanf("%lld",&b[i]), b[i] %= MOD;
    for(int i = 1;i <= n;i++) sum += (n-i+1) * b[i], sum %= MOD;
    for(int i = 1;i <= n;i++){
        ll tmp = ( sum * i ) % MOD;
        for(int j = 1;j < i;j++) tmp = ( tmp - ( ( ( ( b[j] * (n+1) ) % MOD ) * (i-j) ) % MOD ) + MOD ) % MOD;
        ans = ( ans + tmp * a[i] ) % MOD;
    }
    printf("%lld\n",ans);
    return 0;
}

那么如何才能拿到满分呢？考虑把刚刚推出来的柿子进一步优化。观察发现，将柿子中的求和项裂开来，得到:

\[f(m) = m \times sum - (n+1)(m\sum_{i=1}^{m-1}b_i - \sum_{i = 1}^{m - 1} b_i \times i) \]

可以预处理出 \(b_i\) 以及 \(b_i \times i\) 的前缀和，将柿子优化到 \(O(n)\)，于是得到了 \(Accepted\) 的代码：

#include<bits/stdc++.h>
#define MAXN 500010
#define MOD 1000000007ll
using namespace std;
typedef long long ll;
ll n;
ll sum,ans;
ll a[MAXN],b[MAXN],sum_b[MAXN],sum_mult_b[MAXN];
int main(){
    // freopen("sum.in","r",stdin);
    // freopen("sum.out","w",stdout);
    scanf("%lld",&n);
    for(ll i = 1;i <= n;i++) scanf("%lld",&a[i]), a[i] %= MOD;
    for(ll i = 1;i <= n;i++) scanf("%lld",&b[i]), b[i] %= MOD;
    for(ll i = 1;i <= n;i++){
        sum_b[i] = (sum_b[i-1] + b[i]) % MOD;
        sum_mult_b[i] = (sum_mult_b[i-1] + (b[i] * i) % MOD) % MOD;
    }
    for(ll i = 1;i <= n;i++) sum += (n-i+1) * b[i], sum %= MOD;
    for(ll i = 1;i <= n;i++){
        ll tmp = ( (sum * i) % MOD - ( (n+1) * ((( i * sum_b[i-1] ) % MOD - sum_mult_b[i-1] + MOD) % MOD + MOD) % MOD ) % MOD + MOD ) % MOD;
        ans = ( ans + ( tmp * a[i] ) % MOD ) % MOD;
    }
    printf("%lld\n",ans);
    return 0;
}