由于b[i] |n-a[i]

所以n-a[i]为b[i]的倍数,记为y倍

则n-a[i] = y*b[i];

即n+b[i]*y = a[i];

明显转化为欧几里得为ax+by=c; 这里a = 1,x = n,b = b[i],c = a[i];

然后求出x的最小值