SGU 248. Integer Linear Programming( 背包dp )

看了半天...发现就是个背包...然后就不打算敲了. 看了一眼forum..顿时吓傻..其他人用了gcd啊什么的各种奇怪的东西..然后还是敲了个背包结果就AC了= =既然写了代码就扔上来吧...

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#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn = 1000009;

const int INF = 0X3F3F3F3F;

int V[maxn];

int n, N, c[3];

int main() {

scanf("%d", &n);

for(int i = 0; i < n; i++) scanf("%d", c + i);

scanf("%d", &N);

memset(V, INF, sizeof V); V[0] = 0;

for(int i = 0; i < n; i++)

for(int v = c[i]; v <= N; v++)

V[v] = min(V[v], V[v - c[i]] + 1);

if(V[N] != INF)

printf("%d\n", V[N]);

else

puts("-1");

return 0;

}

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248. Integer Linear Programming

time limit per test: 0.25 sec.
memory limit per test: 65536 KB

input: standard
output: standard

You are to solve some problem of integer linear programming. It is posed in the following way. Let x[i] be a variable which is required to be a non-negative integer (for any i from [1..N]). The goal is to minimize the function f(x[1], x[2],..., x[N])=x[1]+x[2]+...+x[N] (objective function) satisfying the constraint c[1]*x[1]+c[2]*x[2]+...+c[N]*x[N]=V.
The point X=(x[1], x[2],..., x[N]) that satisfies the constraint is called "feasible". All feasible points form a feasible set.
To make things clear, let us consider the following example N=2, c[1]=2, c[2]=4, V=6. There are only two feasible points: (1, 1) and (3, 0).
Clearly, the point (1, 1) is the optimal solution, because f(1, 1)<f(3, 0).

Input

The first line of input contains a single positive integer N (0<N<=3). The second line contains N positive integers c[i] separated by whitespaces (0<c[i]<=10^6). The last line contains positive integer V (0<V<=10^6).

Output

On the first line of the output file print the minimal possible value of the function f, or "-1" (without quotes) if the problem has no solution.

Sample test(s)

Input

Test #1
2
2 4
6

Test #2
2
7 4
9

Output

Test #1
2

Test #2
-1

Note

See picture: