Problem:
Score : 300 points

Problem Statement

There are N+1 towns. The i-th town is being attacked by A_i monsters.

We have N heroes. The i-th hero can defeat monsters attacking the i-th or (i+1)-th town, for a total of at most B_i monsters.

What is the maximum total number of monsters the heroes can cooperate to defeat?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N+1}
B_1 B_2 ... B_N

Output

Print the maximum total number of monsters the heroes can defeat.

Sample Input 1

2
3 5 2
4 5

Sample Output 1

9

If the heroes choose the monsters to defeat as follows, they can defeat nine monsters in total, which is the maximum result.

The first hero defeats two monsters attacking the first town and two monsters attacking the second town.

The second hero defeats three monsters attacking the second town and two monsters attacking the third town.

Sample Input 2

3
5 6 3 8
5 100 8

Sample Output 2

22

Sample Input 3

2
100 1 1
1 100

Sample Output 3

3

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void) {
    int n, i;
    scanf("%d", &n);
    int a[n + 1], b[n];
    for ( i = 0; i <= n; i++ ) {
        scanf("%d", &a[i]);
    }
    for ( i = 0; i < n; i++ ) {
        scanf("%d", &b[i]);
    }

    int remain_b = 0, remain_a, ans = 0;
    /* 1 番目の街から n 番目の街までを考える */
    for ( i = 0; i < n; i++ ) {
        /* まず (i - 1) 番目の街の残り戦力をすべて消費 */
        remain_a = a[i] - remain_b;
        /* 残り戦力で足りていたら全部倒す */
        if ( remain_a <= 0 ) {
            ans += a[i];
            remain_b = b[i];
        }
        /* 残り戦力で足りなければその街の勇者が戦う */
        else {
            /* その街の勇者の戦力不足 */
            if ( b[i] <= remain_a ) {
                ans += b[i];
                remain_b = 0;
            }
            /* その街の勇者に余力がある */
            else {
                ans += a[i];
                remain_b = b[i] - remain_a;
            }
        }
    }
    /* (n + 1) 番目の街には専属の勇者がいないので余りで倒せる分だけ足す */
    if ( remain_b > a[n] ) {
        ans += a[n];
    }
    else {
        ans += remain_b;
    }
    printf("%d\n", ans);
    return 0;
}