uoj121 NOI2013—向量内积

Posted 2020-08-30 MashiroSky

http://uoj.ac/problem/121 (题目链接)

题意

　　给出${n}$个${d}$维向量，问是否有两个不同的向量的内积是${k}$的倍数。

Solution

　　又卡了一上午常数，我弃了T_T。

　　右转题解→_→：llg

代码

// uoj121
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<ctime>
#define RG register
#define LL long long
#define inf (1ll<<30)
#define MOD 1000000007
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
inline int gi() {
	int x=0,f=1;char ch=getchar();
	while (ch<\'0\' || ch>\'9\') {if (ch==\'-\') f=-1;ch=getchar();}
	while (ch>=\'0\' && ch<=\'9\') {x=x*10+ch-\'0\';ch=getchar();}
	return x*f;
}

const int maxn=100010,maxd=110;
int a[maxn][maxd],l[maxn],r[maxn];
int u[maxn],v[maxn],p[maxn],tmp[maxn];
int n,d,K,tt;

inline void inc(int &x,RG int y) {
	x+=y;while (x>=K) x-=K;
}
inline int Z(RG int i,RG int j) {
	if (K==2) return a[i][j];
	RG int w=a[i][l[j]]*a[i][r[j]];
	return w>=K ? w-K : w;
}
inline bool solve() {
	for (RG int i=1;i<=n;++i) v[i]=rand()%K,u[i]=v[i],tmp[i]=0;
	
	int sum=0,c=0;
	for (RG int i=1;i<=n;i++) inc(sum,v[i]);
	for (RG int i=1;i<=n;i++) {
		int t=sum-1*v[i]+p[i]*v[i]+K;
		v[i]=0,inc(v[i],t);
	}
	for (RG int i=1;i<=d;i++)
		for (RG int j=1;j<=n;j++) inc(tmp[i],u[j]*Z(j,i));
	if ((double)clock()/CLOCKS_PER_SEC>=4.5) return 0;
	for (RG int i=1;i<=d;i++) u[i]=tmp[i],tmp[i]=0;
	for (RG int i=1;i<=n;i++) {
		for (RG int j=1;j<=d;j++) inc(tmp[i],u[j]*Z(i,j));
		if (tmp[i]!=v[i]) {c=i;break;}
	}

	if (!c) return 0;
	for (RG int i=1;i<=n;i++) if (i!=c) {
			RG int q=0;
			for (RG int j=1;j<=tt;j++) inc(q,a[i][j]*a[c][j]);
			if (!q) {
				printf("%d %d\\n",min(i,c),max(i,c));
				return 1;
			}
		}
	return 1;
}
int main() {
	srand(time(NULL));
	n=gi(),d=gi(),K=gi();
	for (RG int i=1;i<=n;i++)
		for (RG int j=1;j<=d;j++) a[i][j]=gi()%K;
	tt=d;
	if (K==3) {
		d*=d;
		for (RG int i=1;i<=d;i++) l[i]=(i-1)%tt+1,r[i]=(i-l[i])/tt+1;
	}
	for (RG int i=1;i<=n;i++) {
		for (RG int j=1;j<=tt;j++) inc(p[i],a[i][j]);
		if (K==3) (p[i]*=p[i])%=K;
	}
	RG int T=5,flag=0;
	while (T--) {
		flag=solve();
		if (flag) break;
	}
	if (!flag) puts("-1 -1");
	//printf("%.3lf",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}