HDU 5610 Baby Ming and Weight lifting

Problem Description

Baby Ming is fond of weight lifting. He has a barbell pole(the weight of which can be ignored) and two different kinds of barbell disks(the weight of which are respectively $a$ and $b$), the amount of each one being infinite.

Baby Ming prepare to use this two kinds of barbell disks to make up a new one weighted $C$(the barbell must be balanced), he want to know how to do it.

Input

In the first line contains a single positive integer $T$, indicating number of test case.

For each test case:

There are three positive integer $a,b$, and $C$.

$1\le T\le 1000,0<a,b,C\le 1000,a\ne b$

Output

For each test case, if the barbell weighted $C$ can’t be made up, print Impossible.

Otherwise, print two numbers to indicating the numbers of $a$ and $b$ barbell disks are needed. (If there are more than one answer, print the answer with minimum $a+b$)

Sample Input

21 2 61 4 5 

Sample Output

2 2
Impossible
简单枚举就可以了，我还傻乎乎的用了个背包。
#include<string>#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<queue>#include<map>#include<cmath>using namespace std;const int maxn = 1e5 + 10;typedef long long LL;int T, n, m;int a, b, c, f[maxn];int main(){    scanf("%d", &T);    while (T--)    {        scanf("%d%d%d", &a, &b, &c);        if ((c & 1) == 0)        {            c = c / 2;            for (int i = 1; i <= c; i++) f[i] = -1;            f[0] = 0;            for (int i = 0; i <= c; i++)            {                if (f[i] >= 0)                {                    if (f[i + a] == -1) f[i + a] = f[i] + 1;                    else f[i + a] = min(f[i + a], f[i] + 1);                    if (f[i + b] == -1) f[i + b] = f[i] + 1;                    else f[i + b] = min(f[i + b], f[i] + 1);                }            }            if (f[c] >= 0) {                for (int i = 0; i <= f[c]; i++)                {                    if (i*a + (f[c] - i)* b == c)                    {                        printf("%d %d\n", i*2, (f[c] - i)*2);    break;                    }                }            }            else printf("Impossible\n");        }        else printf("Impossible\n");    }    //while (scanf("%d", &n) != EOF){}    return 0;}