Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined)

因为C题卡住了,这次炸了.

我以为模拟肯定T了,所以就写了拓展欧几里得.

然后就被欧几里得坑了.

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用拓展欧几里得做,要把横向和纵向运动分开.

x轴:

f(x)=t%2n,2n−t%2n

y轴:

f(x)=t%2m,2m−t%2m

然后把这两个公式展开一下,在拓展欧几里得求一下就完事了.

(调试出翔)

#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef long long longl;const longl INF=1LL<<60;template <class T> void put(T x) {    if (x==0) {        return;    }    put(x/10);    putchar(x%10^48);}template <class T> void print(T x) {    if (x<0) {        putchar('-'),x=-x;    } else if (x==0) {        putchar('0');        return;    }    put(x);}int exgcd(int a,int b,int &x,int &y) {    if(b==0) {        x=1;y=0;        return a;    }    int r=exgcd(b,a%b,x,y);    longl t=x;x=y;y=t-1LL*a/b*y;    return r;}int N,M;int n,m,k;int x,y,gcd;longl solve(int a,int b) {    if ((b-a)%gcd) return INF;    int t=(b-a)/gcd;    longl k=1LL*x*t;    k=(k%m+m)%m;    return 2LL*N*k+a;}int main() {    int n,m,k;    cin>>n>>m>>k;    N=n,M=m;    n<<=1,m<<=1;    gcd=exgcd(n,m,x,y);    n/=gcd,m/=gcd;    ::n=n,::m=m;    exgcd(n,m,x,y);    for (int a,b,i=1;i<=k;++i) {        longl t=INF;        scanf("%d%d",&a,&b);        t=min(t,solve(a,b));        t=min(t,solve(a,2*M-b));        t=min(t,solve(2*N-a,b));        t=min(t,solve(2*N-a,2*M-b));        if (t==INF) puts("-1");        else print(t),putchar('\n');    }    return 0;}