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....
1/* ***********************************************
2Author :111qqz
3Created Time :Thu 13 Oct 2016 04:23:13 PM CST
4File Name :code/poj/2142.cpp
5************************************************ */
6#include <cstdio>
7#include <cstring>
8#include <iostream>
9#include <algorithm>
10#include <vector>
11#include <queue>
12#include <set>
13#include <map>
14#include <string>
15#include <cmath>
16#include <cstdlib>
17#include <ctime>
18#define fst first
19#define sec second
20#define lson l,m,rt<<1
21#define rson m+1,r,rt<<1|1
22#define ms(a,x) memset(a,x,sizeof(a))
23typedef long long LL;
24#define pi pair < int ,int >
25#define MP make_pair
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30const int inf = 0x3f3f3f3f;
31LL a,b,d;
32LL exgcd( LL a,LL b,LL &x,LL &y)
33{
34 if (b==0)
35 {
36 x = 1;
37 y = 0;
38 return a;
39 }
40 LL ret = exgcd(b,a%b,x,y);
41 LL tmp = x;
42 x = y;
43 y = tmp - a/b*y;
44 return ret;
45}
46LL num ( LL x)
47{
48 if (x<0) return -x;
49 return x;
50}
51LL cal( LL x,LL y)
52{
53 return a*num(x)+b*num(y);
54}
55bool ok( LL x,LL y,LL gx,LL gy)
56{
57 if (num(x)+num(y)>num(x+gx)+num(y-gy)) return true;
58 if (num(x)+num(y)==num(x+gx)+num(y-gy)&&cal(x,y)>cal(x+gx,y-gy)) return true;
59 return false;
60}
61bool ok2( LL x,LL y,LL gx,LL gy)
62{
63 if (num(x) + num(y) > num(x-gx) + num(y+gy)) return true;
64 if (num(x) + num(y) ==num (x-gx) + num(y+gy)&&cal(x,y)>cal(x-gx,y+gy)) return true;
65 return false;
66}
67int main()
68{
69 #ifndef ONLINE_JUDGE
70 freopen("code/in.txt","r",stdin);
71 #endif
72 while (~scanf("%lld%lld%lld",&a,&b,&d))
73 {
74 if (a==0&&b==0&&d==0) break;
75 LL x,y;
76 LL gcd = exgcd(a,b,x,y);
77 x = x * d/gcd;
78 y = y * d/gcd;
79 LL gx = b/gcd;
80 LL gy = a/gcd;
81 while (ok(x,y,gx,gy))
82 {
83 x = x + gx;
84 y = y - gy;
85 }
86 while ( ok2(x,y,gx,gy))
87 {
88 x = x-gx;
89 y = y+gy;
90 }
91 printf("%lld %lld\n",num(x),num(y));
92 }
93 #ifndef ONLINE_JUDGE
94 fclose(stdin);
95 #endif
96 return 0;
97}