CF1005A Tanya and Stairways

题目描述

Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from 1 1 1 to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains 3 3 3 steps, and the second contains 4 4 4 steps, she will pronounce the numbers 1,2,3,1,2,3,4 1, 2, 3, 1, 2, 3, 4 1,2,3,1,2,3,4 .

You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

输入输出格式

输入格式：

The first line contains n n n ( 1≤n≤1000 1 \le n \le 1000 1≤n≤1000 ) — the total number of numbers pronounced by Tanya.

The second line contains integers a1,a2,…,an a_1, a_2, \dots, a_n a1​,a2​,…,an​ ( 1≤ai≤1000 1 \le a_i \le 1000 1≤ai​≤1000 ) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with x x x steps, she will pronounce the numbers 1,2,…,x 1, 2, \dots, x 1,2,…,x in that order.

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

输出格式：

In the first line, output t t t — the number of stairways that Tanya climbed. In the second line, output t t t numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.

输入输出样例

输入样例#1： 复制

7
1 2 3 1 2 3 4

输出样例#1： 复制

2
3 4 

输入样例#2： 复制

4
1 1 1 1

输出样例#2： 复制

4
1 1 1 1 

输入样例#3： 复制

5
1 2 3 4 5

输出样例#3： 复制

1
5 

输入样例#4： 复制

5
1 2 1 2 1

输出样例#4： 复制

3
2 2 1 

很简单，上代码：

#include<cstdio>
using namespace std;
int a[1000],b[1000];
int n,temp,j=0,ans=0;
int main()
{

        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&temp);
             a[temp]++;
             if(temp!=1)
             {
                 b[j]++;

             }
             if(temp==1)
                j++;
        }
        printf("%d\n",a[1]);
        for(int i=1;i<=j;i++)
            printf("%d ",(b[i]+1));
        return 0;

}