zoukankan      html  css  js  c++  java
  • CodeForces

    Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from 11 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 33 steps, and the second contains 44 steps, she will pronounce the numbers 1,2,3,1,2,3,41,2,3,1,2,3,4.

    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.

    Input

    The first line contains nn (1≤n≤10001≤n≤1000) — the total number of numbers pronounced by Tanya.

    The second line contains integers a1,a2,…,ana1,a2,…,an (1≤ai≤10001≤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 xx steps, she will pronounce the numbers 1,2,…,x1,2,…,x in that order.

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

    Output

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

    Examples

    Input

    7
    1 2 3 1 2 3 4
    

    Output

    2
    3 4 

    Input

    4
    1 1 1 1
    

    Output

    4
    1 1 1 1 

    Input

    5
    1 2 3 4 5
    

    Output

    1
    5 

    Input

    5
    1 2 1 2 1
    

    Output

    3
    2 2 1 

    题解:模拟 统计1的个数就是楼梯数 然后记录每个1前面的数就是步骤数。

    代码:

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    
    using namespace std;
    
    int main() {
    
    	int n;
    	cin>>n;
    	int a[1005],b[1005];
    	int cnt=1;
    	cin>>a[0];
    	for(int t=1; t<n; t++) {
    		scanf("%d",&a[t]);
    		if(a[t]==1) {
    			b[cnt]=a[t-1];
    			cnt++;
    		}
    	}
    	cout<<cnt<<endl;
    	b[cnt]=a[n-1];
    	cnt++;
    	for(int t=1; t<cnt; t++) {
    		cout<<b[t]<<" ";
    	}
    	return 0;
    }
  • 相关阅读:
    程序员的最大挑战
    12个有效的提高编程技能的方法
    风雨20年:我所积累的20条编程经验
    java的继承机制
    Java中获得程序当前路径的4中方法
    关于String的hashCode
    使用三目运算符时注意的一个问题
    linux查找符合条件的文件并删除
    Tomcat性能优化及JVM内存工作原理
    Linux(Centos)下调整分区大小(以home和根分区为例)
  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781906.html
Copyright © 2011-2022 走看看