UVALive - 6428（扩展欧几里德）
http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=48388
前段时间偶然碰到的一道题，今天突然想到没把它记录下来。
比较不错的扩展欧几里德求解的应用
题意求是否满足ax+by=c gcd(a,b）==1 a>=0&&b>=0
数比较大 在求解时LL会溢出，用JAVA改写了一遍程序。
扩展欧几里德可以求出一组满足条件的解p= p1+b/gcd(a,b)*t q = q1+a/gcd(a,b)*t （t为任意整数）
因为上面的解有条件 所以根据p>=0 q>=0可以确定出来t的范围 在范围内枚举t可以求出解来 判断是否满足条件即可。
注意一下特殊情况，为0的情况。
1 import java.text.*; 2 import java.io.*; 3 import java.util.*; 4 import java.math.*; 5 import java.applet.*; 6 public class Main 7 { 8 static BigInteger x,y,o = BigInteger.valueOf(0),o1 = BigInteger.valueOf(1); 9 static BigInteger gcd(BigInteger a,BigInteger b) 10 { 11 if(b.compareTo(BigInteger.valueOf(0))==0) 12 return a; 13 else 14 return gcd(b,a.mod(b)); 15 } 16 static BigInteger exgcd(BigInteger a,BigInteger b) 17 { 18 if(b.compareTo(o)==0) 19 { 20 x = o1; y = o; return a; 21 } 22 BigInteger d = exgcd(b,a.mod(b)); 23 BigInteger temp = x; 24 x = y; 25 y = temp.subtract(a.divide(b).multiply(y)) ; 26 return d; 27 } 28 public static void main(String[] args) 29 { 30 Scanner cin = new Scanner(System.in); 31 BigInteger a,b,c; 32 while(cin.hasNext()) 33 { 34 a = cin.nextBigInteger(); 35 b = cin.nextBigInteger(); 36 c = cin.nextBigInteger(); 37 if(c.compareTo(a)<0&&c.compareTo(b)<0) 38 { 39 System.out.println("NO"); 40 continue; 41 } 42 BigInteger o = BigInteger.valueOf(0); 43 if(a.compareTo(o)==0||b.compareTo(o)==0) 44 { 45 if(a.compareTo(o)==0&&b.compareTo(o)==0) 46 { 47 if(c.compareTo(o)==0) 48 System.out.println("YES"); 49 else 50 System.out.println("NO"); 51 } 52 else if(a.compareTo(o)==0) 53 { 54 if(c.mod(b).compareTo(o)==0) 55 System.out.println("YES"); 56 else 57 System.out.println("NO"); 58 } 59 else 60 { 61 if(c.mod(a).compareTo(o)==0) 62 System.out.println("YES"); 63 else 64 System.out.println("NO"); 65 } 66 continue; 67 } 68 if(c.compareTo(b)==0||c.compareTo(a)==0) 69 { 70 System.out.println("YES"); 71 continue; 72 } 73 BigInteger t = gcd(a,b); 74 if(c.mod(t).compareTo(o)!=0) 75 { 76 System.out.println("NO"); 77 continue; 78 } 79 exgcd(a,b); 80 x = x.multiply(c.divide(t)); 81 y = y.multiply(c.divide(t)); 82 BigInteger k1 = b.divide(t); 83 BigInteger k2 = a.divide(t); 84 int flag = 0; 85 BigInteger tx=x,ty=y; 86 BigInteger f1 = (x.multiply(BigInteger.valueOf(-1))).divide(k1).subtract(BigInteger.valueOf(1)); 87 BigInteger f2 = y.divide(k2).add(BigInteger.valueOf(1)); 88 BigInteger e = f1; 89 while(e.compareTo(f2)<=0) 90 { 91 tx = k1.multiply(e).add(x); 92 ty = k2.multiply(e); 93 ty = y.subtract(ty); 94 if(tx.compareTo(o1)>=0&&ty.compareTo(o1)>=0) 95 { 96 if(gcd(tx,ty).compareTo(BigInteger.valueOf(1))==0) 97 { 98 //System.out.println(tx+" "+ty); 99 flag = 1; 100 break; 101 } 102 } 103 e = e.add(BigInteger.valueOf(1)); 104 } 105 if(flag==1) 106 System.out.println("YES"); 107 else 108 System.out.println("NO"); 109 } 110 } 111 }
