​http://poj.org/problem?id=2142​

先求ax+by=gcd(a,b)的一组特解(x0 y0) 而后的ax+by=c的通解(x0*(c/gcd)+t*(b/gcd) y0*(c/gcd)-t*(a/gcd))

题目要求的是|x|+|y|最小 其次|a*x|+|b*y|最小的解

看别人博客知道 当c>0时 若a>b则x非负 反之则y非负 这样只要保证a>b则|x0*(c/gcd)+t*(b/gcd)|单增 而|y0*(c/gcd)-t*(a/gcd)|先减后增 又因为a>b 所以减速大于增速 所以|x0*(c/gcd)+t*(b/gcd)|+|y0*(c/gcd)-t*(a/gcd)|也是先减后增的 且在当y0*(c/gcd)-t*(a/gcd)=0时取最小值 这样整除得t 前后枚举一下即可

 

满足|x|+|y|最小的解只有两三个左右 再从里边挑|a*x|+|b*y|最小的解即可

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll N=0x3f3f3f3f3f3f3f3f;

void exgcd(ll a,ll b,ll& x,ll& y)
{
ll tx,ty;
if(b==0){
x=1,y=0;
return;
}
exgcd(b,a%b,tx,ty);
x=ty,y=tx-a/b*ty;
}

ll getabs(ll val)
{
if(val>=0) return val;
else return -val;
}

int main()
{
ll a,b,d,x,y,gcd,p,i,minn1,minn2,xx,yy,ansx,ansy;
bool flag;
while(scanf("%lld%lld%lld",&a,&b,&d)!=EOF){
if(a==0&&b==0&&d==0) break;
if(a<b){
swap(a,b);
flag=0;
}
else flag=1;
exgcd(a,b,x,y);
//printf("*%lld %lld*\n",x,y);
gcd=a*x+b*y;
if(d%gcd!=0) printf("no solution\n");
else{
p=y*d/a,minn1=N,minn2=N;
//printf("***%lld %lld***\n",d*x/gcd,d*y/gcd);
//printf("*%lld*\n",p);
for(i=p-5;i<=p+5;i++){
//printf("*%lld %lld*\n",(b*d*i)/(gcd*gcd),(a*d*i)/(gcd*gcd));
xx=d*x/gcd+(b*i)/gcd;
yy=d*y/gcd-(a*i)/gcd;
xx=getabs(xx),yy=getabs(yy);
if(minn1>xx+yy){
minn1=xx+yy,minn2=a*xx+b*yy;
ansx=xx,ansy=yy;
}
else if(minn1==xx+yy){
if(minn2>a*xx+b*yy){
minn2=a*xx+b*yy;
ansx=xx,ansy=yy;
}
}
}
if(flag) printf("%lld %lld\n",ansx,ansy);
else printf("%lld %lld\n",ansy,ansx);
}
}
return 0;
}

//3 1 10000