poj 2142 The Balance (扩展欧几里得算法)
题意：给出a,b,d，分别表示a,b两种刻度的砝码，以及要称量的物体重量为d.现在保证能称量出给定重量的物体，问两种砝码个数的和最小的时候，两种砝码分别有多少。如果有多组解，那么要求weight of(ax + by) 最小。
思路：求特解直接扩展欧几里得...
关键是怎么找到绝对值和最小的。。
我就是两个方向跑了下。。。
一开始因为把weight of (ax+by) （求得还是绝对值最小）理解成了 ax+by最小。。导致WA了半天。。。。sigh....
/* ***********************************************
Author :111qqz
Created Time :Thu 13 Oct 2016 04:23:13 PM CST
File Name :code/poj/2142.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
LL a,b,d;
LL exgcd( LL a,LL b,LL &x,LL &y)
{
if (b==0)
{
x = 1;
y = 0;
return a;
}
LL ret = exgcd(b,a%b,x,y);
LL tmp = x;
x = y;
y = tmp - a/b*y;
return ret;
}
LL num ( LL x)
{
if (x<0) return -x;
return x;
}
LL cal( LL x,LL y)
{
return a*num(x)+b*num(y);
}
bool ok( LL x,LL y,LL gx,LL gy)
{
if (num(x)+num(y)>num(x+gx)+num(y-gy)) return true;
if (num(x)+num(y)==num(x+gx)+num(y-gy)&&cal(x,y)>cal(x+gx,y-gy)) return true;
return false;
}
bool ok2( LL x,LL y,LL gx,LL gy)
{
if (num(x) + num(y) > num(x-gx) + num(y+gy)) return true;
if (num(x) + num(y) ==num (x-gx) + num(y+gy)&&cal(x,y)>cal(x-gx,y+gy)) return true;
return false;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("code/in.txt","r",stdin);
#endif
while (~scanf("%lld%lld%lld",&a,&b,&d))
{
if (a==0&&b==0&&d==0) break;
LL x,y;
LL gcd = exgcd(a,b,x,y);
x = x * d/gcd;
y = y * d/gcd;
LL gx = b/gcd;
LL gy = a/gcd;
while (ok(x,y,gx,gy))
{
x = x + gx;
y = y - gy;
}
while ( ok2(x,y,gx,gy))
{
x = x-gx;
y = y+gy;
}
printf("%lld %lld\n",num(x),num(y));
}
#ifndef ONLINE_JUDGE
fclose(stdin);
#endif
return 0;
}