Codeforces Round #325 (Div. 2) A. Alena's Schedule 暴力枚举 字符串

A. Alena's Schedule

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Alena has successfully passed the entrance exams to the university and is now looking forward to start studying.

One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes).

The University works in such a way that every day it holds exactly n lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks).

The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the n pairs she knows if there will be a class at that time or not.

Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university.

Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home.

Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 100) — the number of lessons at the university.

The second line contains n numbers ai (0 ≤ ai ≤ 1). Number ai equals 0, if Alena doesn't have the i-th pairs, otherwise it is equal to 1. Numbers a1, a2, ..., an are separated by spaces.

Output

Print a single number — the number of pairs during which Alena stays at the university.

Sample test(s)

input

5
0 1 0 1 1

output

4

input

7
1 0 1 0 0 1 0

output

4

input

1
0

output

0

Note

In the first sample Alena stays at the university from the second to the fifth pair, inclusive, during the third pair she will be it the university waiting for the next pair.

In the last sample Alena doesn't have a single pair, so she spends all the time at home.

题目分析：原始题意见题目，转化后：转化后给你一连串的01字符串，求其中单独的1和处于101状态的0出现的次数，，转化忽的题意是看了别人的才想到的，自己的又是非常笨拙的暴力

先贴上好的代码

 #include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int a[105];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int cnt=0;
        for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
                if(a[i]==1)
                    cnt++;
            }
         for(int i=1;i<=n-2;i++)
            if(a[i]==1&&a[i+1]==0&&a[i+2]==1)
              cnt++;
         printf("%d\n",cnt);
    }
    return 0;
}

　　再贴上自己的挫的代码，水题也要得进步

#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = ;
const double eps = 1e-;
const int inf = 0x3f3f3f3f;
int a[];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int s=,e=;
        for(int i=;i<=n;i++)
            {
                scanf("%d",&a[i]);
                if(a[i]&&i<s)
                    s=i;
                if(a[i]&&i>e)
                    e=i;
            }
        if(s==)
        {
            printf("0\n");
            continue;
        }
        int cnt=e-s+;
        for(int i=s;i<=e;)
            {
                int j=i;
                if(a[i]==)
                {
                while(a[i]==)
                     i++;
                if(i-j>=)
                    cnt-=(i-j);
                }
                else i++;
            }
         printf("%d\n",cnt);
    }
    return ;
}