Problem:
Score : 300 points

Problem Statement

There is an integer sequence A of length N whose values are unknown.

Given is an integer sequence B of length N-1 which is known to satisfy the following:

B_i \geq \max(A_i, A_{i+1})

Find the maximum possible sum of the elements of A.

Constraints

All values in input are integers.

2 \leq N \leq 100

0 \leq B_i \leq 10^5

Input

Input is given from Standard Input in the following format:

N
B_1 B_2 ... B_{N-1}

Output

Print the maximum possible sum of the elements of A.

Sample Input 1

3
2 5

Sample Output 1

9

A can be, for example, ( 2 , 1 , 5 ), ( -1 , -2 , -3 ), or ( 2 , 2 , 5 ). Among those candidates, A = ( 2 , 2 , 5 ) has the maximum possible sum.

Sample Input 2

2
3

Sample Output 2

6

Sample Input 3

6
0 153 10 10 23

Sample Output 3

53

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

#include <stdio.h>

int main() {
    int n;
    int a[105], b[105],minlist[105];
    scanf("%d",&n);
    for (int i = 0; i < 105; ++i) {
        a[i] = -1;
        b[i] = -1;
    }
    for (int i = 0; i < n-1; ++i) {
        scanf("%d",&b[i]);
        minlist[i] = b[i];
    }

    //MINLIST TIMEEE
    int z,j;
    for (int i = 0; i < n-1; ++i) {
        for (j = i + 1; j < n; ++j) {
            if (minlist[i] > minlist[j])  {
                z = minlist[i];
                minlist[i] = minlist[j];
                minlist[j] = z;
            }
        }
     }
    /*for (int i = 0; i < n-1; ++i) {
         printf("%d\n", minlist[i]);
    }*/
    for (int i = 0; i < n-1; ++i) {
        for (int j = 0; j < n-1; ++j) {
            if (minlist[i] == b[j]) {
                if ((a[j] > b[j])||(a[j] == -1)) {
                    a[j] = b[j];
                    //printf("a[%d]: %d\n",j,a[j]);
                }
                //printf("a[%d+1]: %d\n",j,a[j+1]);
                if ((a[j+1] > b[j])||(a[j+1] == -1)) {
                    a[j+1] = b[j];
                    //printf("a[%d+1]: %d\n",j,a[j+1]);
                }
            }
        }
    }
    int summ = 0;
    for (int i = 0; i < n; ++i) {
        summ = summ + a[i];
        //printf("%d\n", a[i]);
    }
    printf("%d\n",summ);


    return 0;
}