Codeforces 366C C. Dima and Salad (DP) __codeforces

Topic Link:

Codeforces 366C topic to the effect:

Give n items, there are two attributes, ask the last property is the sum of the second attribute of K times, the first property is the maximum number. Topic Analysis: We will make a deformation, weight for a[i]-b[i]*k, the benefit is a, then we only need to do the weight of the backpack, the quality of negative to take absolute value to do a backpack, and then the weight of the same backpack benefits and is the current weight of the maximum income.
AC Code:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define MAX 107

using namespace std;

int dp [MAX * MAX], dd [MAX * MAX];

int n, k, a [MAX], b [MAX], c [MAX];

int main ()
{
    while (~ scanf ("% d% d", & n, & k))
    {
        for (int i = 0; i <n; i ++)
            scanf ("% d", & a [i]);
        for (int i = 0; i <n; i ++)
            scanf ("% d", & b [i]);
        for (int i = 0; i <n; i ++)
            c [i] = a [i] -b [i] * k;
        memset (dp, -0x3f, sizeof (dp));
        memset (dd, -0x3f, sizeof (dd));
        dp [0] = dd [0] = 0;
        for (int i = 0; i <n; i ++)
            if (c [i]> = 0)
                for (int j = 10000; j> = c [i]; j--)
                    dp [j] = max (dp [j-c [i]] + a [i], dp [j]);
        for (int i = 0; i <n; i ++)
            if (c [i] <0)
            {
                c [i] = -c [i];
                for (int j = 10000; j> = c [i]; j--)
                    dd [j] = max (dd [j-c [i]] + a [i], dd [j]);
            }
        int ans = -1;
        for (int i = 0; i <= 10000; i ++)
        {
            if (dd [i] == 0 && dp [i] == 0)
                continue;
            ans = max (ans, dp [i] + dd [i]);
        }
        printf ("% d \ n", ans);
    }
}