Codeforces Round #214 (Div. 2)??Dima and Salad_html/css_WEB-ITnose

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Question meaning:
One row a[i], one row b[i], a and b are in one-to-one correspondence. Select any number pair such that sigma(a)/sigma(b) is equal to k, and find the maximum value of sigma(a) at this time

Analysis:
The key to this question is to compare sigma(a)/ Processing of sigma(b)==k. For this formula, it is obviously not possible to use the ratio of each number, because it cannot be accumulated; and it is a double type, so it is impossible to DP
consider the impact of each number pair on this formula. If each number are all a = k * b, then it is obviously possible; if a is less than k * b, then in the whole, the current number with fewer pairs must have some number pairs to compensate, that is, remember x = k * b - a, the sum of x for all selected pairs should be zero.
Then, each number pair actually becomes a number that can be positive or negative. Find the maximum value of sigma (b) of several numbers whose sum is zero, separate the positive and negative, and use a backpack to solve it

const int MAXN = 110;const int M = 210000;int tastes[MAXN], calories[MAXN];int v[MAXN];int dp[2][M];int main(){//    freopen("in.txt", "r", stdin);    int n, m;    while (~RII(n, m))    {        REP(i, n) RI(tastes[i]);        REP(i, n) RI(calories[i]);        REP(i, n) v[i] = tastes[i] - calories[i] * m;        CLR(dp, -INF);        dp[0][0] = dp[1][0] = 0;        REP(i, n)        {            int cnt = v[i] < 0, val = abs(v[i]);            FED(j, M - val - 1, 0)                dp[cnt][j + val] = max(dp[cnt][j + val], dp[cnt][j] + tastes[i]);        }        int ans = 0;        REP(i, M)            ans = max(ans, dp[0][i] + dp[1][i]);        if (ans <= 0)            puts("-1");        else            WI(ans);    }    return 0;}