Problem:
Score : 200 points

Problem Statement

There are N gems. The value of the i-th gem is V_i.

You will choose some of these gems, possibly all or none, and get them.

However, you need to pay a cost of C_i to get the i-th gem.

Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.

Find the maximum possible value of X-Y.

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq C_i, V_i \leq 50

Input

Input is given from Standard Input in the following format:

N
V_1 V_2 ... V_N
C_1 C_2 ... C_N

Output

Print the maximum possible value of X-Y.

Sample Input 1

3
10 2 5
6 3 4

Sample Output 1

5

If we choose the first and third gems, X = 10 + 5 = 15 and Y = 6 + 4 = 10.
We have X-Y = 5 here, which is the maximum possible value.

Sample Input 2

4
13 21 6 19
11 30 6 15

Sample Output 2

6

Sample Input 3

1
1
50

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <numeric>
#include <string.h>
using namespace std;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = 1<<29;
const long long  INFL = 1e18;
const int MOD = 1000000007;
const int MAX = 210000;
int N;
vector<int>C, V;
void solve() {
    cin >> N;
    C.resize(N);
    V.resize(N);
    for (int i = 0; i < N; i++) {
        cin >> V[i];
    }
    for (int i = 0; i < N; i++) {
        cin >> C[i];
    }
    if (N == 1) {
        if (V[0] - C[0] > 0) {
            cout << V[0] - C[0] << endl;
            return;
        } else {
            cout << 0 << endl;
            return;
        }
    }
    long long ans = 0;
    long long sum_v = 0;
    long long sum_c = 0;
    for (int i = 0; i < N; i++) {
        long long tmp = 0;
        sum_v = 0;
        sum_c = 0;
        for (int j = i; j < N; j++) {
            tmp = (sum_v + V[j]) - (sum_c + C[j]);
            if (tmp > ans) {
                sum_v += V[j];
                sum_c += C[j];
                ans = (sum_v - sum_c);
            }
        }
    }
    if (ans > 0) {
        cout << ans << endl;
    } else {
        cout << 0 << endl;
    }
    return;
}
//逆元を求める関数
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++) {
      fac[i] = fac[i-1] * i % MOD;
      inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
      finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
//組み合わせ
long long COM (int n, int k) {
    if(n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n-k] % MOD) % MOD;
}
int main(int args, char* argv[]) {
    solve();
}