poj2891（中国剩余定理）

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//满足x%a1=r1,x%a2=r2,..x%an%rn,求x//因为a1,a2..不互质，不能用剩余定理//只能一步一步计算/*关键：x=((r/d*x)%t+t)%t;t=a2/d；r1=x*a1+r1;a1=a1*a2/d;*/#include"stdio.h"#include"string.h"typedef __int64 LL;LL x,y,d;void exgcd(LL a,LL b){if(b==0){x=1;y=0;d=a;}else {exgcd(b,a%b);LL t=x;x=y;y=t-(a/b)*y;}}int main(){LL n;LL a1,a2;LL r1,r2;LL flag,i;while(scanf("%I64d",&n)!=-1){scanf("%I64d%I64d",&a1,&r1);flag=0;for(i=0;i<n-1;i++){scanf("%I64d%I64d",&a2,&r2);LL r=r2-r1;exgcd(a1,a2);if(r%d!=0){flag=1;continue;}LL t=a2/d;x=((r/d*x)%t+t)%t;r1=x*a1+r1;a1=a1*a2/d;}if(flag==1)printf("-1\n");else printf("%I64d\n",r1);}return 0;}