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    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;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781906.html
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