[ZLXOI2015]殉国 数论 扩展欧几里得

题目大意：已知a,b,c，求满足ax+by=c （x>=0,y>=0）的（x+y）最大值与最小值与解的个数。

直接exgcd，求出x，y分别为最小正整数的解，然后一算就出来啦

#include<cstdio>#include<iostream>#define ll long longusing namespace std;ll a,b,c,x,y,d,bd,ad,X1,Y1,X2,Y2;ll Abs(ll x){return x>=0?x:-x;}ll exgcd(ll a,ll b,ll &x,ll &y){if(b==0){x=1; y=0;return a;}int gcd=exgcd(b,a%b,x,y);int t=x;x=y;y=t-(a/b)*x;return gcd;}int main(){//freopen("BlackHawk.in","r",stdin);//freopen("BlackHawk.out","w",stdout);scanf("%lld%lld%lld",&a,&b,&c);d=exgcd(a,b,x,y);//printf("%lld\n",d);if(c%d!=0){       printf("-1 -1\n0\n");       return 0;}x*=c/d; y*=c/d;//printf("%lld  %lld\n",x,y);bd=b/d; ad=a/d;X1=(x%bd+bd)%bd;ll t=(X1-x)/bd;Y1=y-t*ad;ll num1=X1+Y1;//printf("%lld  %lld\n",X1,Y1);if(Y1<0){       printf("-1 -1\n0\n");       return 0;}Y2=(y%ad+ad)%ad;t=(Y2-y)/ad;X2=x-t*bd;ll num2=X2+Y2;//printf("%lld  %lld\n",X2,Y2);if(X2<0){       printf("-1 -1\n0\n");       return 0;}if(num2<num1){t=num1; num1=num2; num2=t;}printf("%lld %lld\n",num1,num2);t=Abs((X1-X2)/bd)+1;printf("%lld\n",t);}