Problem:
 Score : 400 points    Problem Statement Dolphin is planning to generate a small amount of a certain chemical substance C. In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b . He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy. The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock. The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan). Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C. Find the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.      Constraints  1≦N≦40   1≦a_i,b_i≦10   1≦c_i≦100   1≦M_a,M_b≦10   gcd(M_a,M_b)=1  a_i , b_i , c_i , M_a and M_b are integers.         Input The input is given from Standard Input in the following format:  N  M_a  M_b  a_1  b_1  c_1  a_2  b_2  c_2  :  a_N  b_N  c_N       Output Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print -1 instead.        Sample Input 1 3 1 1
1 2 1
2 1 2
3 3 10      Sample Output 1 3  The amount of money spent will be minimized by purchasing the packages of chemicals 1 and 2 . In this case, the mixture of the purchased chemicals will contain 3 grams of the substance A and 3 grams of the substance B, which are in the desired ratio: 3:3=1:1 . The total price of these packages is 3 yen.       Sample Input 2 1 1 10
10 10 10      Sample Output 2 -1  The ratio 1:10 of the two substances A and B cannot be satisfied by purchasing any combination of the packages. Thus, the output should be -1 .

Solution (Python):

import sys,copy,math,heapq,itertools as it,fractions,re,bisect,collections as coll

inf = 1e10
def gcd(a, b):
    while b: a, b = b, a%b
    return a

N, Ma, Mb = map(int, raw_input().split())
abc = [map(int, raw_input().split()) for i in xrange(N)]

dp = [[0]*405 for i in xrange(405)]

for a, b, c in abc:
    for i in xrange(404-a, -1, -1):
        for j in xrange(404-b, -1, -1):
            if dp[i][j]:
                if dp[i + a][j + b] == 0: dp[i + a][j + b] = dp[i][j] + c
                else: dp[i + a][j + b] = min(dp[i + a][j + b], dp[i][j] + c)
    if dp[a][b] == 0: dp[a][b] = c
    else: dp[a][b] = min(dp[a][b], c)

ans = inf
for i in xrange(1, 405):
    for j in xrange(1, 405):
        g = gcd(i, j)
        if i/g == Ma and j/g == Mb and dp[i][j]:
            ans = min(ans, dp[i][j])

print ans if ans < inf else -1

Solution (Python):

N, Ma, Mb = [int(_) for _ in input().split()]
T = [[int(_) for _ in input().split()] for i in range(N)]

dp = [[float('inf') for i in range(401)] for i in range(401)]
dp[0][0] = 0

sum_a = 0
sum_b = 0
for x in range(N):
    a, b, c = T[x]
    sum_a += a
    sum_b += b
    for i in list(range(a, sum_a+1))[::-1]:
        for j in list(range(b, sum_b+1))[::-1]:
            dp[i][j] = min(dp[i][j], dp[i-a][j-b]+c)

ans = float('inf')
for i in range(1, sum_a+1):
    for j in range(1, sum_b+1):
        if i * Mb == j * Ma:
            ans = min(ans, dp[i][j])

print(ans if ans != float('inf') else -1)

Solution (Python):
INF = 1010101010

def main():
    N,Ma,Mb = map(int,input().split())

    a = [0 for _ in range(N)]
    b = [0 for _ in range(N)]
    c = [0 for _ in range(N)]

    for i in range(N):
        a[i],b[i],c[i] = map(int,input().split())

    dp = [[[INF for k in range(401)] for j in range(401)] for i in range(41)]
    dp[0][0][0] = 0

    for i in range(N):
        for A in range(401):
            for B in range(401):
                dp[i+1][A][B] = min(dp[i+1][A][B], dp[i][A][B])
                if A+a[i]<=400 and B+b[i]<=400:
                    dp[i+1][A+a[i]][B+b[i]] = min(dp[i+1][A+a[i]][B+b[i]], dp[i][A][B]+c[i])

    ans = INF
    for A in range(1,401):
        for B in range(1,401):
            if Ma*B == Mb*A:
                ans = min(ans, dp[N][A][B])

    return -1 if ans == INF else ans

if __name__ == '__main__':
    print(main())

Solution (Python):
import sys

stdin = sys.stdin
def na(): return map(int, stdin.readline().split())
def ns(): return stdin.readline().strip()


n,ma,mb = na()
abcs = []
for i in range(n):
    abcs.append(list(na()))
I = 9999999999
dp = [[I]*420 for _ in range(420)]
dp[0][0] = 0
sa = 1; sb = 1
for abc in abcs:
    for i in range(sa-1,-1,-1):
        for j in range(sb-1,-1,-1):
            dp[i+abc[0]][j+abc[1]] = min(dp[i+abc[0]][j+abc[1]], dp[i][j] + abc[2])
    sa += abc[0]
    sb += abc[1]
x = ma
y = mb
ret = I
while x < 420 and y < 420:
    ret = min(ret, dp[x][y])
    x += ma
    y += mb
if ret == I:
    print(-1)
else:
    print(ret)