下发长为 (d) 的 (01) 串 (pi)，输入长为 (d) 的 (01) 串 (sigma)，求 (d(pi,sigma)=sum_{i=1}^d|pi_i-sigma_i|)。
(d=2^{19})，代码长度限制 (2 ext{KB})，2-approximate。
数理基础题，我自闭。
JL Lemma
给定 (epsilonin(0,1)) 和 (m) 个点 (x_1,cdots,x_minR^N)，(forall n>log m/(epsilon^2/2-epsilon^3/3))，存在线性映射 (f:R^N ightarrowR^n)，使得
[left|frac{|f(x_i)-f(x_j)|^2}{|x_i-x_j|^2}-1
ight|leepsilon
]
这实际上是一个超级著名的结论，因为它在将数据量压缩到对数阶的同时保相当程度的线性性。然而不太会证
JL Lemma 的特例
设 (mathbf v) 是 (n) 维向量，(A) 在 ({-1,1}^{d imes n}) 中均匀随机，则 (mathbb E[|Amathbf v|^2]=d|mathbf v|^2, ext{Var}[|Amathbf v|^2]=O(|v|^4))。
证明有手就行？
于是我们固定随机数种子，随机一个 (A)。算出 (|A(pi-sigma)|^2) 然后 (/d) 四舍五入即可。
先把 (Acdotpi) 算出来打个表，然后计算时把 (Acdotsigma) 算出来即可。
取 (n=2^8)，时间复杂度 (O(frac{nd}w))，代码长度复杂度 (O(n))。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1<<19, B = 256, biao[B] = {-171,345,-491,-15,-463,-335,487,-429,-257,-1161,-289,-545,-341,-273,-323,455,-621,25,921,-865,57,-7,131,705,-133,-513,715,-411,497,-745,765,-1105,305,247,571,-297,47,-661,-951,-371,197,447,-859,-31,41,701,371,243,435,677,571,-509,-885,891,-337,-879,-811,337,-1075,309,-433,873,-285,-441,-979,-373,969,-279,155,375,-95,225,-187,-187,-227,-427,229,-585,869,489,341,-311,5,27,187,1,-629,281,593,207,-175,645,693,-321,-485,-529,-163,-493,623,859,165,-287,439,113,-653,591,-73,417,-37,-719,307,491,537,141,619,-155,-389,-381,673,-613,-187,-637,-45,367,-289,-827,605,-127,-1059,93,629,-279,-89,353,-13,-1231,341,609,-219,-125,-813,205,-655,607,-217,-91,-431,227,-75,-275,407,69,429,107,-49,-725,-357,29,-531,681,-143,-41,-407,835,-533,471,-85,419,51,-51,-1095,-53,717,-175,-681,-663,-73,-641,1281,189,99,519,-365,-505,269,335,-1101,-727,193,-331,289,-149,409,43,65,-3,1319,121,-881,-271,-431,-123,1167,249,-801,-109,501,-41,-469,81,-27,-115,959,-373,649,-701,-267,-583,-171,-1123,-75,-87,-525,119,-709,111,869,165,-381,-829,561,15,743,-11,-123,233,-83,145,-271,67,841,133,-229,-421,-809,123,149,-877,-21,119,-821,-531,-449,263,125,675};
char str[N+5]; unsigned S[N>>5]; int sum; LL ans;
int main(){
mt19937 rng(20210325); scanf("%s", str);
for(int i = 16383;~i;-- i){
S[i] = strtol(str+(i<<5), NULL, 2) ^ rng(); str[i<<5] = 0;
sum += __builtin_popcount(S[i]);
} for(int _ = 0;_ < B;++ _){ int res = 0;
for(int i = 0;i < 16384;++ i) res += __builtin_popcount(rng() & S[i]);
res = res * 2 - sum - biao[_]; ans += (LL)res * res;
} printf("%lld
", (ans + 128) / B);
}