标签：bar 方差 求和 na sum len int 51nod1687 1n
题目
解题思路
简化公式
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\sum_{i=1}^n(a_i-\bar a)^2\\=\sum_{i=1}^n(a_i^2-2a_i\bar a+\bar a^2)\\=\sum_{i=1}^na_i^2-2\bar a\sum_{i=1}^nai+n\bar a^2\\=\sum_{i=1}^na_i^2-2n\bar a^2+n\bar a^2\\=\sum_{i=1}^na_i^2-n\bar a^2
∑i=1n(ai−aˉ)2=∑i=1n(ai2−2aiaˉ+aˉ2)=∑i=1nai2−2aˉ∑i=1nai+naˉ2=∑i=1nai2−2naˉ2+naˉ2=∑i=1nai2−naˉ2
考虑增量法
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n\bar a^2=\frac{(\sum_{i=1}^na_i)^2}{n}
naˉ2=n(∑i=1nai)2
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\frac{(x + a)^2}{n+1}-\frac{x^2}{n}\\=\frac{nx^2+2nxa+na^2-(n+1)x^2}{n(n+1)}\\=\frac{2nax+na^2-x^2}{n(n+1)}\\=\frac{2ax+a^2}{n+1}-\frac{x^2}{n(n+1)}
n+1(x+a)2−nx2=n(n+1)nx2+2nxa+na2−(n+1)x2=n(n+1)2nax+na2−x2=n+12ax+a2−n(n+1)x2
无法摆脱与长度的关系，不可行。
考虑枚举长度，同时计算长度相同的区间
我们知道区间长度和左端点L就可以计算右端点R了。
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ans=\sum_{len=1}^n\sum_{L=1}^{n-len+1}(\sum_{i=L}^Ra_i^2-len\bar a^2)
ans=∑len=1n∑L=1n−len+1(∑i=LRai2−lenaˉ2)
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\sum_{len=1}^n\sum_{L=1}^{n-len+1}\sum_{i=L}^Ra_i^2=\sum_{i=1}^na_i^2*i*(n-i+1)
∑len=1n∑L=1n−len+1∑i=LRai2=∑i=1nai2∗i∗(n−i+1)可以简单求出。
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\sum_{len=1}^n\sum_{L=1}^{n-len+1}len\bar a^2
∑len=1n∑L=1n−len+1lenaˉ2
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=\sum_{len=1}^n\frac{\sum_{L=1}^{n-len+1}(\sum_{i=L}^Ra_i)^2}{len}
=∑len=1nlen∑L=1n−len+1(∑i=LRai)2
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=\sum_{len=1}^n\frac{\sum_{L=1}^{n-len+1}(\sum_{i=L}^Ra_i^2+2\sum_{i=L}^R\sum_{j=i+1}^Ra_ia_j)}{len}
=∑len=1nlen∑L=1n−len+1(∑i=LRai2+2∑i=LR∑j=i+1Raiaj)
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\sum_{len=1}^n\frac{\sum_{L=1}^{n-len+1}\sum_{i=L}^Ra_i^2}{len}
∑len=1nlen∑L=1n−len+1∑i=LRai2也能够简单维护。
令
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f(len)=\sum_{L=1}^{n-len+1}\sum_{i=L}^R\sum_{j=i+1}^Ra_ia_j
f(len)=∑L=1n−len+1∑i=LR∑j=i+1Raiaj
我们发现
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f(len)
f(len)的二阶差分可以分为两部分，一部分可以用NTT维护，一部分可以简单维护。
整体复杂度
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O(Tnlog_2n)
O(Tnlog2n)。
代码
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
void read(int &x) {
x = 0; char c = getchar();
while (c < '0' || c > '9') c = getchar();
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
}
const int N = 3e5 + 100;
namespace NTT {
#define pw(n) (1<<n)
const int N = 262144, P = 998244353, g = 3;//或P=1004535809
int n, m, bit, bitnum = 0, a[N + 5], b[N + 5], rev[N + 5];
void getrev(int l) {
for (int i = 0; i < pw(l); i++) {
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
}
}
int fastpow(int a, int b) {
int ans = 1;
for (; b; b >>= 1, a = 1LL * a*a%P) {
if (b & 1)ans = 1LL * ans*a%P;
}
return ans;
}
void NTT(int *s, int op) {
for (int i = 0; i < bit; i++)if (i < rev[i])swap(s[i], s[rev[i]]);
for (int i = 1; i < bit; i <<= 1) {
int w = fastpow(g, (P - 1) / (i << 1));
for (int p = i << 1, j = 0; j < bit; j += p) {
int wk = 1;
for (int k = j; k < i + j; k++, wk = 1LL * wk*w%P) {
int x = s[k], y = 1LL * s[k + i] * wk%P;
s[k] = (x + y) % P;
s[k + i] = (x - y + P) % P;
}
}
}
if (op == -1) {
reverse(s + 1, s + bit);
int inv = fastpow(bit, P - 2);
for (int i = 0; i < bit; i++)s[i] = 1LL * s[i] * inv%P;
}
}
int solve(int *aa, int nn, int *bb, int mm, int *c) {
n = nn; m = mm;
bit = bitnum = 0;
for (int i = 0; i <= n; i++) a[i] = aa[i];
for (int i = 0; i <= m; i++) b[i] = bb[i];
m += n;
for (bit = 1; bit <= m; bit <<= 1)bitnum++;
getrev(bitnum);
NTT(a, 1);
NTT(b, 1);
for (int i = 0; i < bit; i++) a[i] = 1LL * a[i] * b[i] % P;
NTT(a, -1);
for (int i = 0; i < bit; i++) c[i] = a[i];
for (int i = 0; i < bit; i++) a[i] = b[i] = 0;
return bit;
}
}
int n;
int sa[N], sum[N], pre[N], suf[N], aa[N], bb[N];
int main() {
//freopen("0.txt", "r", stdin);
int T; read(T);
while (T--) {
read(n);
for (int i = 1; i <= n; i++) read(sa[i]);
double ans = 0;
for (int i = 1; i <= n; i++) {
ans += 1LL * i * (n - i + 1) * sa[i] * sa[i];
sum[i] = sum[i - 1] + sa[i] * sa[i];
pre[i] = pre[i - 1] + sa[i];
}
for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + sa[i];
ll r = sum[n];
for (int len = 1; len <= n; len++) {
ans -= r / double(len);
int L = len + 1, R = n - len;
if (L <= r) r += sum[R] - sum[L - 1];
else r -= sum[L] - sum[R - 1];
}
for (int i = 1; i <= n; i++) {
aa[i] = sa[i];
bb[i] = sa[n - i + 1];
}
NTT::solve(aa, n + 1, bb, n + 1, aa);
ll r1 = 0, r2 = 0;
for (int len = 2; len <= n; len++) {
r1 += aa[n - len + 2] - sa[n - len + 2] * suf[n - len + 3] - sa[len - 1] * pre[len - 2];
r2 += r1;
ans -= 2 * r2 / double(len);
}
printf("%.2lf\n", ans);
for (int i = 0; i <= n + 1; i++) sum[i] = pre[i] = suf[i] = 0;
}
return 0;
}
标签：bar,方差,求和,na,sum,len,int,51nod1687,1n 来源： https://blog.csdn.net/qq_39586831/article/details/114992127
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