POJ - 1061 扩展gcd

题意:求\((n-m)t+Lk=x-y\)的解\(t\)
#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<string>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int maxn = 1e6+11;

const int oo = 0x3f3f3f3f;

const double eps = 1e-7;

typedef long long ll;

ll read(){

    ll x=0,f=1;register 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;

}

ll exgcd(ll a,ll b,ll &x,ll &y){

	if(b==0){

		x=1;y=0;

		return a;

	}else{

		ll gcd=exgcd(b,a%b,x,y);

		ll tmp=x;

		x=y;y=tmp-a/b*x;

		return gcd;

	}

}

int main(){

	int kase=0;

	ll x,y,m,n,L,MOD;

	while(cin>>x>>y>>m>>n>>L){

		ll t,k;

		ll gcd=exgcd(n-m+L,L,t,k);

		if((x-y+L)%gcd!=0) cout<<"Impossible"<<endl;

		else{

			t=t*(x-y+L)/gcd;

			MOD=L/gcd;

			t=(t%MOD+MOD)%MOD;

			cout<<t<<endl;

		}

	}

	return 0;

}
巴特西

POJ - 1061 扩展gcd

最新文章

热门文章
