方程ax+by+c=0; 问有多少对x,y. 满足 x1<=x<=x2, y1<=y<=y2ax+by=-c. c=-c,后， 扩展欧几里得求一下，x,y;x,y满足 ax+by=gcd(a,b);x*=c/gcd(a,b);y*=c/gcd(a,b);后x,y满足 ax+by=c;通解：x0=x+k*b/gcd(a,b);y0=y-k*
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