BZOJ 2299 向量(裴蜀定理)

时间：2024-11-10 12:38:08

题意:给你一对数a,b，你可以任意使用(a,b), (a,-b), (-a,b), (-a,-b), (b,a), (b,-a), (-b,a), (-b,-a)这些向量，问你能不能拼出另一个向量(x,y)。

实际上前四个向量能拼出(ma,nb)(m%2=n%2).后四个向量拼出(xb,ya)(x%2=y%2).

这样可以枚举这四个未知数在模二意义下的解。这两个向量相加为(ma+xb,nb+ya).

对于ma+xb=X.根据系数的奇偶性，如果有系数为奇数，可使得等式两边都减去一个数使得系数都为偶数，这样再同除以二。

就是一般的用裴蜀定理来判断这类方程是否有解的过程了。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

const int N=;

//Code begin...

bool check(LL x, LL y, LL gcd){return x%gcd==&&y%gcd==;}

int main ()

{

    int T;

    LL a, b, x, y, gcd;

    scanf("%d",&T);

    while (T--) {

        scanf("%lld%lld%lld%lld",&a,&b,&x,&y);

        if (a==&&b==) {puts(x==&&y==?"Y":"N"); continue;}

        if (a==||b==) gcd=(a==?b:a);

        else gcd=__gcd(a,b);

        gcd*=;

        if (check(x,y,gcd)||check(x-a,y-b,gcd)||check(x-b,y-a,gcd)||check(x-a-b,y-a-b,gcd)) puts("Y");

        else puts("N");

    }

    return ;

}

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