欧几里德

ax+by=gcd(a,b)与ax+by=c 模版

LL ext_gcd(LL a,LL b,LL& x,LL& y){ //解 ax+by=gcd(a,b);
    LL t,ret;
    if (!b){
        x=1,y=0;
        return a;
    }
    ret=ext_gcd(b,a%b,x,y);
    t=x,x=y,y=t-a/b*y;
    return ret;
}
LL ext(LL a,LL b,LL c){  //解 ax+by=c 
    LL x, y;
    LL gcd=ext_gcd(a,b,x,y);
    if(c%gcd) {  //当c不为gcd(a,b)的倍数时，即无整数解
        cout<<"failed"<<endl;   
        return 0;
    }
    x*=c/gcd; y*=c/gcd;
    LL dx=b/gcd,dy=a/gcd;   //x=x+i*dx ,y=y-i*dy; 
//求 x>=0&& y>=0时 i可取的范围 [low,up] ；
    LL low=ceil(-x/(double)dx);   
    LL up=floor(y/(double)dy);

    if(low>up) cout<<"failed"<<endl;  //当 low>up (即上界大于下界）不存在整数解
}

 

uva 10104 - Euclid Problem

#include<iostream>
#include<cstring>
#include <cstdio>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include <set>
#include<ctime>
#include<cmath>
#include <cstdlib>
#include<algorithm>
using namespace std;
#define LL long long 
LL ext_gcd(LL a,LL b,LL& x,LL& y){     //该算法得到的 x 和 y 即是绝对值和最小的
    LL t,ret;
    if (!b){
        x= 1,y= 0;
        return a;
    }
    ret=ext_gcd(b,a%b,x,y);
    t=x,x=y,y=t-a/b*y;
    return ret;
}
int main(){
    LL a,b;
    while(scanf("%lld%lld",&a,&b)!=EOF){
        LL x,y;
        LL gcd=ext_gcd(a,b,x,y);
        printf("%lld %lld %lld\n",x,y,gcd);
    }
}

 

uva 10090 - Marbles

#include<iostream>
 #include<cstring>
 #include <cstdio>
 #include<string>
 #include<queue>
 #include<vector>
 #include<map>
 #include <set>
 #include<hash_set>
 #include<ctime>
 #include<cmath>
 #include <cstdlib>
 #include<algorithm>
 using namespace std;
 #define LL long long 
 #define nMAX (1<<10+10)
 #define mMAX 110
 LL n,n1,n2,c1,c2;
 LL ext_gcd(LL a,LL b,LL& x,LL& y){ //解 ax+by=gcd(a,b);
     LL t,ret;
     if (!b){
         x=1,y=0;
         return a;
     }
     ret=ext_gcd(b,a%b,x,y);
     t=x,x=y,y=t-a/b*y;
     return ret;
 }
 LL ext(LL a,LL b,LL c){  //解 ax+by=c 
     LL x, y;
     LL gcd=ext_gcd(a,b,x,y);
     if(n%gcd) {
         cout<<"failed"<<endl; 
         return 0;
     }
     x*=c/gcd; y*=c/gcd;
     LL dx=b/gcd,dy=a/gcd;
     LL low=ceil(-x/(double)dx);
     LL up=floor(y/(double)dy);
 
     if(low>up) cout<<"failed"<<endl; 
     else {
         if(c1*n2<=c2*n1){
             x=x+up*dx;
             y=y-up*dy;
         }else {
             x=x+low*dx;
             y=y-low*dy;
         }
         cout<<x<<" "<<y<<endl;
     }
     return gcd;
 }
 int main(){
     while(cin>>n&&n){
         cin>>c1>>n1>>c2>>n2;
         ext(n1,n2,n);
     }
 }

 

zoj 3593 One Person Game 

#include<iostream>
#include<cstring>
#include <cstdio>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include <set>
#include<ctime>
#include<cmath>
#include <cstdlib>
#include<algorithm>
#include <iomanip>
#include <bitset>
using namespace std;
#define LL  long long 

const LL MAX = 1<<25;
const LL nMAX =50;
const LL INF =(((1LL)<<63)-1);
const LL MOD= 1000000007;
LL ans;
LL ext_gcd(LL a,LL b,LL& x,LL& y){ 
    LL t,ret;
    if (!b){
        x=1,y=0;
        return a;
    }
    ret=ext_gcd(b,a%b,x,y);
    t=x,x=y,y=t-a/b*y;
    return ret;
}
LL ext(LL a,LL b,LL c){  
    LL x,y;
    LL gcd=ext_gcd(a,b,x,y);
    if(c%gcd) {  
        return 0;
    }
    x*=c/gcd; y*=c/gcd;
    LL dx=b/gcd,dy=a/gcd;   //x=x+i*dx ,y=y-i*dx; 
    //求 x>=0&& y>=0时 i可取的范围 [low,up] ；
    LL low=floor(-x/(double)dx);   
    LL up=ceil(y/(double)dy);
    ans=min((LL)(abs(x)+abs(y)),ans);
    ans=min((LL)(abs(x+low*dx)+abs(y-low*dy)),ans);
    ans=min((LL)(abs(x+up*dx)+abs(y-up*dy)),ans);
    
    low=ceil(-x/(double)dx);   
    up=floor(y/(double)dy);
    ans=min((LL)(abs(x)+abs(y)),ans);
    ans=min((LL)(abs(x+low*dx)+abs(y-low*dy)),ans);
    ans=min((LL)(abs(x+up*dx)+abs(y-up*dy)),ans);
    return 1;
}
int main(){
    LL A,B,a,b,T;
    scanf("%lld",&T);
    while(T--){
        scanf("%lld%lld%lld%lld",&A,&B,&a,&b);
        if(A>B) swap(A,B);
        ans=INF;
        ext(a,b,B-A);
        ext(a+b,b,B-A);
        ext(a+b,a,B-A);
        if(ans!=INF) printf("%lld\n",ans);
        else printf("-1\n");
    }
}