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POJ 2142 The Balance

2016-05-05 20:57 295 查看
题目链接:http://poj.org/problem?id=2142

题意:给两种砝码a,b,和待称量物品的重量d,用x个a和y个b来将物品称出,并使得x+y最小。

思路:a * x + b * y = d,扩展欧几里得。先让x为最小非负整数解,算出y;再让y为最小非负整数解,算出x,两种情况取最小的即可。

#include <cstdio>

#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;

#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)

#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod %100000007

int a,b,d;

int exgcd(int a,int b,int &x,int &y)
{
if ( b == 0 )
{
x = 1;
y = 0;
return a;
}
int r = exgcd( b , a % b , x , y );
int t = x;
x = y;
y = t - a/b * y;
return r;
}

void slove(int a,int b,int d)
{
int x,y;
int gcd = exgcd( a , b , x , y );
x*=d/gcd;
y*=d/gcd;
int addx = b/gcd;
int addy = a/gcd;

int m1,m2;
int ansx,ansy;
m1 = m2 = inf;

x = ( x % addx + addx )% addx;
y = (d - a*x)/b;
ansx = x;
ansy = abs(y);
m1 = x + abs(y);
m2 = a*x + b*abs(y);

y = ( y % addy + addy )% addy;
x = (d - b*y)/a;

if ( abs(x) + abs(y) < m1 )
{
m1 = abs(x) + abs(y);
ansx = abs(x);
ansy = abs(y);
}
else if ( abs(x) + abs(y) == m1 && a*abs(x) + b*abs(y) < m2 )
{
ansx = abs(x);
ansy = abs(y);
}

printf("%d %d\n",ansx,ansy);
}

int main()
{
while( ~scanf("%d%d%d",&a,&b,&d) )
{
if ( a + b + d == 0 ) break;
slove(a,b,d);
}
return 0;
}
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