VJ【背包】

/*DescriptionDima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to have a walk, Inna decided to cook something.Dima and Seryozha have n fruits in the fridge. Each fruit has two parameters: the taste and the number of calories. Inna decided to make a fruit salad, so she wants to take some fruits from the fridge for it. Inna follows a certain principle as she chooses the fruits: the total taste to the total calories ratio of the chosen fruits must equal k. In other words,  , where aj is the taste of the j-th chosen fruit and bj is its calories.Inna hasn't chosen the fruits yet, she is thinking: what is the maximum taste of the chosen fruits if she strictly follows her principle? Help Inna solve this culinary problem — now the happiness of a young couple is in your hands!Inna loves Dima very much so she wants to make the salad from at least one fruit.InputThe first line of the input contains two integers n, k(1?≤?n?≤?100,?1?≤?k?≤?10). The second line of the input contains n integers a1,?a2,?...,?an(1?≤?ai?≤?100) — the fruits' tastes. The third line of the input contains n integers b1,?b2,?...,?bn(1?≤?bi?≤?100) — the fruits' calories. Fruit number i has taste ai and calories bi.OutputIf there is no way Inna can choose the fruits for the salad, print in the single line number -1. Otherwise, print a single integer — the maximum possible sum of the taste values of the chosen fruits.Sample InputInput3 210 8 12 7 1Output18Input5 34 4 4 4 42 2 2 2 2Output-1HintIn the first test sample we can get the total taste of the fruits equal to 18 if we choose fruit number 1 and fruit number 2, then the total calories will equal 9. The condition  fulfills, that's exactly what Inna wants.In the second test sample we cannot choose the fruits so as to follow Inna's principle.*/#include<stdio.h>#include<string.h>#define INF 0x3f3f3f3fint a[110], b[110], dp[30010];int max(int x, int y){return x>y?x:y;}int main(){int i, j, k, n;while(scanf("%d%d", &n, &k) != EOF){for(i = 1; i <= n; ++i)scanf("%d", &a[i]);for(i = 1; i <= n; ++i){scanf("%d", &b[i]);b[i] = a[i] - k*b[i];}memset(dp, -INF, sizeof(dp));dp[10000] = 0;for(i = 1; i <= n; ++i){if(b[i] >= 0){for(j = 0; j >= b[i]; --j)dp[j] = max(dp[j], dp[j-b[i]] + a[i]); }else{for(j = 0; j <= 20000; ++j)dp[j] = max(dp[j], dp[j - b[i]] + a[i]);}} if(dp[10000] == 0)dp[10000] = -1;printf("%d\n", dp[10000]);}return 0;}//题意：给出n个a[i] 和n个b[i] (1<=i<=n),从中选出m对数使得 a数列的和 / b数列的和 = k，并且要求a数列的和尽可能的大//思路：根据sum(a[i]) - k*sum(b[i]) = 0 (1<=i<=n) ,我们可以将其看作是一个背包问题，a[i]当作是价值，a[i]-k*b[i]当作是容量//即当容量刚好为0（即背包刚好装满）的时候所能放下的最大价值，由于物品容量可能为负数，所以我们需要扩展背包的容量以便计算//当物品容量>=0时，就是正常的01背包，而当物品容量<0时，我们需要反向循环构成一个逆向的01背包 ，而中间的dp[10000]则作为2个循环背包刚好装满时的最大值