标签:unique require 表达式 mil nim top with mat bsp
Marbles
Input: standard input
Output: standard output
I have some (say, n) marbles (small glass balls) and I am going to buy some boxes to store them. The boxes are of two types:
Type 1: each box costs c1 Taka and can hold exactly n1 marbles
Type 2: each box costs c2 Taka and can hold exactly n2 marbles
I want each of the used boxes to be filled to its capacity and also to minimize the total cost of buying them. Since I find it difficult for me to figure out how to distribute my marbles among the boxes, I seek your help. I want your program to be efficient also.
Input
The input file may contain multiple test cases. Each test case begins with a line containing the integer n (1 <= n <= 2,000,000,000). The second line contains c1and n1, and the third line contains c2 and n2. Here, c1, c2, n1and n2 are all positive integers having values smaller than 2,000,000,000.
A test case containing a zero forn in the first line terminates the input.
Output
For each test case in the input print a line containing the minimum cost solution (two nonnegative integers m1 and m2, where mi= number of Type i boxes required) if one exists, print "failed" otherwise.
If a solution exists, you may assume that it is unique.
Sample Input
43
Sample Output
13 1
题意:一个人有n个弹球。如今要把这些弹球所有装进盒子里。第一种盒子每一个盒子c1美元,能够恰好装n1个弹球。另外一种盒子每一个盒子c2元。能够恰好装n2个弹球。找出一种方法把这n个弹球装进盒子,每一个盒子都装满,而且花费最少的钱。
分析:如果第一种盒子买m1个,另外一种盒子买m2个,则n1*m1 + n2*m2 = n。由扩展欧几里得 ax+by=gcd(a,b)= g,如果n%g!=0。则方程无解。
联立两个方程。能够解出m1=nx/g, m2=ny/g,所以通解为m1=nx/g + bk/g, m2=ny/g - ak/g,
又由于m1和m2不能是负数,所以m1>=0, m2>=0,所以k的范围是 -nx/b <= k <= ny/a。且k必须是整数。
如果
k1=ceil(-nx/b)
k2=floor(ny/b)
假设k1>k2的话则k就没有一个可行的解。于是也是无解的情况。
设花费为cost,则cost = c1*m1 + c2*m2,
把m1和m2的表达式代入得
cost=c1*(-xn/g+bk/g)+c2*(yn/g-ak/g) = ((b*c1-a*c2)/g)*k+(c1*x*n+c2*y*n)/g
这是关于k的一次函数。单调性由b*c1-a*c2决定。
若b*c1-a*c2 >= 0,k取最小值(k1)时花费最少;否则,k取最大值(k2)时花费最少。
#include<iostream> #include<cmath> using namespace std; typedef long long LL; LL extend_gcd(LL a, LL b, LL *x, LL *y) { LL xx, yy, g; if(a < b) return extend_gcd(b, a, y, x); if(b == 0) { *x = 1; *y = 0; return a; } else { g = extend_gcd(b, a%b, &xx, &yy); *x = yy; *y = (xx - a/b*yy); return g; } } int main() { LL n, c1, n1, c2, n2, x, y; while(cin >> n && n) { cin >> c1 >> n1 >> c2 >> n2; LL g = extend_gcd(n1, n2, &x, &y); if(n % g != 0) { cout << "failed" << endl; continue; } LL mink = ceil(-n * x / (double)n2); LL maxk = floor(n*y / (double)n1); // mink <= k <= maxk if(mink > maxk) { cout << "failed" << endl; continue; } if(c1 * n2 <= c2 * n1) { x = n2 / g * maxk + n / g * x; y = n / g * y - n1 / g * maxk; } else { x = n2 / g * mink + n / g * x; y = n / g * y - n1 / g * mink; } cout << x << " " << y << endl; } return 0; }
标签:unique require 表达式 mil nim top with mat bsp
原文地址:http://www.cnblogs.com/cxchanpin/p/6727214.html