模板题_NOIp2012同余方程

扩展gcd的模板题，代码如下

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 #define lowbit(a) ((a)&(-a))
 7 #define max(a, b) ((a)>(b)?(a):(b))
 8 #define min(a, b) ((a)<(b)?(a):(b))
 9 #define MAXN 100010
10 #define PI 3.1415926
11 #define E 2.718281828
12 #define INF 0x777777f
13 typedef long long LL;
14
15 inline int exGcd(int m,int n,int &x,int &y){
16     int x0=1,y0=0,x1=0,y1=1,r=m%n,q=(m-r)/n;x=0;y=1;
17     while(r){x=x0-q*x1;y=y0-q*y1;x0=x1;y0=y1;x1=x;y1=y;m=n;n=r;r=m%n;q=(m-r)/n;} return n;
18 }
19 int main(){
20     int a, b, x, y;
21     scanf("%d%d", &a, &b);
22     exGcd(a, b, x, y);
23     printf("%d", (x+b)%b);
24     return 0;
25 }