好久没有写博了……今天水一水中国生育腚理不互质的情况 设n1与n2不互质
其中gcd(n1,n2)|(a2-a1)才有解的原因如下：
设gcd(n1,n2)=d
n1=d*t1
n2=d*t2
∴x=k1*t1*d+a1
=k2*t2*d+a2
∴k1t1d+a1=k2t2d+a2
∴(k1t1-k2t2)d=a2-a1
∴gcd(n1,n2)|(a2-a1)
例题：poj2891 http://poj.org/problem?id=2891
1 #include<cmath> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<iostream> 6 #include<algorithm> 7 #define eps (0.5) 8 #define RG register 9 using namespace std; 10 typedef long long LL; 11 LL d,k,x,y,hold,lastyu,lastmod,a[10000000],r[10000000]; 12 inline LL gll(){ 13 RG LL x=0;RG bool flag=0;RG char c=getchar(); 14 while((c<'0'||c>'9')&&c!='-') c=getchar(); 15 if(c=='-') c=getchar(),flag=1; 16 while(c>='0'&&c<='9') x=x*10+c-'0',c=getchar(); 17 return flag?-x:x; 18 } 19 inline void e_gcd(RG LL a,RG LL b,RG LL &x,RG LL &y){ 20 if(b==0){d=a;x=1;y=0;return;} 21 e_gcd(b,a%b,x,y); 22 hold=x;x=y;y=hold-a/b*y; 23 } 24 inline LL fbc(RG LL a,RG LL b,RG LL mod){return (a*b-(LL)((double)a/mod*(double)b+eps)*mod+mod)%mod;} 25 inline LL China(){ 26 for (RG int i=1;i<=k;++i){ 27 e_gcd(lastmod,a[i],x,y); 28 if((r[i]-lastyu)%d) return -1; 29 LL gcd=d; 30 e_gcd(lastmod/gcd,a[i]/gcd,x,y); 31 LL now=fbc((r[i]-lastyu)/gcd,x,a[i]/gcd); 32 lastyu=(lastyu+fbc(now,lastmod,lastmod/gcd*a[i]))%(lastmod/gcd*a[i]); 33 lastmod=lastmod/gcd*a[i]; 34 } 35 if(lastyu<0) lastyu+=lastmod; 36 return lastyu; 37 } 38 inline void work(){ 39 for (RG LL i=1;i<=k;++i){ 40 a[i]=gll(); 41 r[i]=gll(); 42 } 43 lastyu=r[1];lastmod=a[1]; 44 printf("%lld ",China()); 45 } 46 int main(){ 47 while(scanf("%lld",&k)!=EOF) work(); 48 return 0; 49 }