Description
You are asked to help her by calculating how many weights are required.
Input
The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.
Output
- You can measure dmg using x many amg weights and y many bmg weights.
- The total number of weights (x + y) is the smallest among those pairs of nonnegative integers satisfying the previous condition.
- The total mass of weights (ax + by) is the smallest among those pairs of nonnegative integers satisfying the previous two conditions.
No extra characters (e.g. extra spaces) should appear in the output.
Sample Input
700 300 200 500 200 300 500 200 500 275 110 330 275 110 385 648 375 4002 3 1 10000 0 0 0
Sample Output
1 3 1 1 1 0 0 3 1 1 49 74 3333 1
解題思路
拓展歐幾里得求最小正解,比較難以理解的是求出一組解後,如何求最小正解的過程,此題再次再次求教了userluoxuan,在此鳴謝
AC代碼
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int INF = 999999999;
int gcd(int a, int b)
{
return b ? gcd(b, a % b) : a;
}
int extend_gcd (int a, int b, int &x, int &y)
{
if(b == 0)
{
x = 1;
y = 0;
return a;
}
int r = extend_gcd(b, a % b, x, y);
int temp = x;
x = y;
y = temp - a / b * y;
return r;
}
int main()
{
int a, b, n, d, x, y, min, min_x, min_y;
while(scanf("%d%d%d", &a, &b, &n) && ( a || b || n) )
{
bool flag = false;
min = INF;
if(a < b)
{
swap(a, b);
flag = true; //原方程等式爲a * x + b * y = n
}
d = gcd(a, b); //求a,b的最大公約數,a,b化成互質的
a /= d;
b /= d;
extend_gcd(a, b, x, y); //求出a * x + b * y = n / gcd(a , b) 的一組解
x *= n / d; //化成原方程的解
y *= n / d;
int t = y / a; //由通解公式有 y(n) = y - a * t ,令y(n) = 0,有 t = y / a,此t爲0點左右的
for(int i = t - 1; i <= t + 1; i++) //查找t左右兩解,尋找最小正解
{
int x_1 = x + b * i;
int y_1 = y - a * i;
if(abs(x_1) + abs(y_1) < min)
{
min = abs(x_1) + abs(y_1);
min_x = abs(x_1);
min_y = abs(y_1);
}
}
if(flag)
printf("%d %d\n", min_y, min_x);
else
printf("%d %d\n", min_x, min_y);
}
return 0;
}