思路浅析:
这题是01背包的一个变形,只要把a[i]作为价值,a[ i ] - k * b[ i ]作为代价,打一个01背包即可
/* CF366C Dima ans Salad */
#include <bits/stdc++.h>
const int N = 1e5;
int dp[110][N * 2 + 10];
int a[110], b[110], v[110];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while (!isdigit(ch))
f = (ch == '-') ? -1 : 1, ch = getchar();
while (isdigit(ch))
x = x * 10 + (ch - '0'), ch = getchar();
return x * f;
}
int main()
{
int n, k;
n = read(), k = read();
for (int i = 1; i <= n; i++)
a[i] = read();
for (int i = 1; i <= n; i++)
b[i] = read(), v[i] = a[i] - k * b[i];
for (int i = 0; i <= N * 2 + 10; i++)
dp[0][i] = -(1 << 30);
dp[0][N] = 0;
for (int i = 1; i <= n; i++)
for (int j = 2 * N; j >= 0; j--)
dp[i][j] = std::max(dp[i - 1][j], dp[i - 1][j - v[i]] + a[i]);
if (dp[n][N] == 0)
printf("-1\n");
else
printf("%d\n", dp[n][N]);
return 0;
}