Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 4781 | Accepted: 2092 |
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
题意是给出了a,b,d的重量。问使用a、b怎么测出d的重量,假设是能够测出的前提下。输出|x|+|y|的最小值。
a*x+b*y=d
之后求一下x的最小值时y的值。再求一遍y的最小值时x的值。两两比较即可。
代码:
#include <iostream> #include <vector> #include <string> #include <cstring> #include <algorithm> #include <cmath> using namespace std; int xx,yy,yue; int a,b,d; vector <int> x_value; vector <int> y_value; void ex_gcd(int a,int b, int &xx,int &yy) { if(b==0) { xx=1; yy=0; yue=a; } else { ex_gcd(b,a%b,xx,yy); int t=xx; xx=yy; yy=t-(a/b)*yy; } } void cal() { int i; int min=abs(x_value[0])+abs(y_value[0]); int min_x=abs(x_value[0]),min_y=abs(y_value[0]); for(i=1;i<2;i++) { if(abs(x_value[i])+abs(y_value[i])==min) { if((abs(x_value[i])*a+abs(y_value[i])*b)<(min_x*a+min_y*b)) { min_x=abs(x_value[i]); min_y=abs(y_value[i]); } } if(abs(x_value[i])+abs(y_value[i])<min) { min=abs(x_value[i])+abs(y_value[i]); min_x=abs(x_value[i]); min_y=abs(y_value[i]); } } cout<<min_x<<" "<<min_y<<endl; } int main() { while(cin>>a>>b>>d) { if(a==0 && b==0 && d==0) break; ex_gcd(a,b,xx,yy); x_value.clear(); y_value.clear(); xx=xx*(d/yue); yy=yy*(d/yue); int r=a/yue; yy=(yy%r+r)%r; int xx0,yy0=yy; xx0=(d-yy*b)/a; x_value.push_back(xx0); y_value.push_back(yy); r=b/yue; xx=(xx%r+r)%r; x_value.push_back(xx); yy=(d-xx*a)/b; y_value.push_back(yy); cal(); } return 0; }
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