【NOI2013】向量内积【随机化】

传送门

题意：给 $n$ 个 $d$ 维向量，询问是否有两个向量内积（对应位乘积和）为 $k$ 的倍数

$\leq100000,d\leq100,k=2,3$

考虑每个向量能否与之前的某一个匹配

如果我们找到某一个与之前的可以匹配，就可以 $O (n d)$ 得到答案。我们要做的是排除不能匹配的答案。

（以下 $m$ 为题中给的 $k$ ）

即

$∀1≤i<n,∑k=1dai,kan,k≠0(modm)\forall1\leq i<n,\sum_{k=1}^{d}a_{i,k}a_{n,k}\neq0\pmod{m}$

当 $m = 2$ 时

$∀1≤i<n,∑k=1dai,kan,k≡1(mod2)\forall1\leq i<n,\sum_{k=1}^{d}a_{i,k}a_{n,k}\equiv1\pmod{2}$

弱化得

$∑i=1n−1∑k=1dai,kan,k≡n−1(mod2)\sum_{i=1}^{n-1}\sum_{k=1}^{d}a_{i,k}a_{n,k}\equiv n-1\pmod{2}$

$∑k=1d(∑i=1n−1ai,k)an,k≡n−1(mod2)\sum_{k=1}^{d}(\sum_{i=1}^{n-1}a_{i,k})a_{n,k}\equiv n-1\pmod{2}$

维护个前缀和判一下，如果不满足说明一定有答案

感性理解，理论上这个答案是随便找得到的，所以随机打乱几次能大概率出解

当 $m = 3$ 时同理

$∀1≤i<n,∑k=1dai,kan,k≡1or2(mod3)\forall1\leq i<n,\sum_{k=1}^{d}a_{i,k}a_{n,k}\equiv1 or 2\pmod{3}$

平方一下

$∀1≤i<n,(∑k=1dai,kan,k)2≡1(mod3)\forall1\leq i<n,(\sum_{k=1}^{d}a_{i,k}a_{n,k})^2\equiv1 \pmod{3}$

$∑i=1n−1(∑k=1dai,kan,k)2≡n−1(mod3)\sum_{i=1}^{n-1}(\sum_{k=1}^{d}a_{i,k}a_{n,k})^2\equiv n-1\pmod{3}$

强行拆开

$∑i=1n−1∑x=1d∑y=1dai,xan,xai,yan,y≡n−1(mod3)\sum_{i=1}^{n-1}\sum_{x=1}^{d}\sum_{y=1}^da_{i,x}a_{n,x}a_{i,y}a_{n,y}\equiv n-1\pmod{3}$

$∑x=1d∑y=1d(∑i=1n−1ai,xai,y)an,xan,y≡n−1(mod3)\sum_{x=1}^{d}\sum_{y=1}^d(\sum_{i=1}^{n-1}a_{i,x}a_{i,y})a_{n,x}a_{n,y}\equiv n-1\pmod{3}$

然后就可以维护了

复杂度 $O(nd^{k-1})$

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#define MAXN 100005
#define MAXM 105
using namespace std;
inline int read()
{int ans=0;char c=getchar();while (!isdigit(c)) c=getchar();while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();return ans;
}
int id[MAXN],a[MAXN][MAXM];
int c[MAXM][MAXM],s[MAXM];
int n,d,k;
inline bool check(int x,int y)
{int sum=0;for (int i=1;i<=d;i++) sum+=a[x][i]*a[y][i];return sum%k==0;
}
int main()
{n=read(),d=read(),k=read();for (int i=1;i<=n;i++)for (int j=1;j<=d;j++)a[i][j]=read()%k;for (int i=1;i<=n;i++) id[i]=i;int T=10;while (T--){random_shuffle(id+1,id+n+1);if (k==2){for (int i=1;i<=d;i++) s[i]=0;for (int i=1;i<=n;i++){int sum=0;for (int j=1;j<=d;j++) sum+=s[j]*a[id[i]][j];if (sum%2!=(i-1)%2){for (int x=1;x<i;x++)if (check(id[x],id[i])){if (id[i]>id[x]) swap(id[i],id[x]);printf("%d %d\n",id[i],id[x]);return 0;}}for (int j=1;j<=d;j++) s[j]+=a[id[i]][j];}}else{for (int i=1;i<=d;i++)for (int j=1;j<=d;j++)c[i][j]=0;for (int i=1;i<=n;i++){int sum=0;for (int x=1;x<=d;x++)for (int y=1;y<=d;y++)sum+=c[x][y]*a[id[i]][x]*a[id[i]][y];if (sum%3!=(i-1)%3){for (int j=1;j<i;j++)if (check(id[j],id[i])){if (id[j]>id[i]) swap(id[j],id[i]);printf("%d %d\n",id[j],id[i]);return 0;}}for (int x=1;x<=d;x++)for (int y=1;y<=d;y++)c[x][y]+=a[id[i]][x]*a[id[i]][y];}}}puts("-1");return 0;
}