[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 long
using 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);
}