The 14th Lanqiao Cup Competition Software Competition Final C/C++ University Group B Test Question D: Merging Sequences

[Lanqiao Cup 2023 Country B] Merged Sequence

【Problem Description】

Xiao Ming found that there are many ways to split a large positive integer into the sum of several positive integers. He adopted two of the solutions and listed them as two arrays { a 1 , a 2 , ⋯ an } \{a_1, a_2, \cdots a_n\}{ a1​,a2​,⋯an​} 和 { b 1 , b 2 , ⋯ b m } \{b_1, b_2, \cdots b_m\} { b1​,b2​,⋯bm​} . The sum of both arrays is the same.

Defining a merge operation can merge two adjacent numbers in an array into a new number. The value of the new number is the sum of the original two numbers. Xiao Ming wants to make the two arrays exactly the same through several merge operations, that is, $n=m$ and for any subscript$i$ full foot$a_i=b_i$. Please calculate the minimum number of merge operations required to complete Xiao Ming's goal.

【Input format】

Enter a total of $3$ lines.
The first line is two positive integers$n,m$ .
The second line$n$ integers a 1 , a 2 , ⋯ , an a_1, a_2, \cdots, a_nseparated by spaces$a_1,a_2,\cdots ,a_n$.
The third line is $m$ integers b 1 , b 2 , ⋯ , bm b_1, b_2, \cdots, b_mseparated by spaces$b_1,b_2,\cdots ,b_m$。

Output format

Output total $1$ line, an integer.

[Sample input]

4 3
1 2 3 4
1 5 4

[Sample output]

1

[Sample description]

Just change $a_2$Japanese $a_3$Merge, array $a$ becomes{ 1 , 5 , 4 } \{1,5,4\}{ 1,5,4 } , that is, and$b$ is the same.

[Evaluation use case scale and agreement]

• For $20\%$ of the data ensures$n,m\le 10^3$。
• For $100\%$ data, guaranteed$n,m\le 10^5$，$0<a_i,b_i\le 10^5$。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    
    
	list<int> a,b;
	int n,m;cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
    
    
		int x;cin>>x;
		a.push_back(x);
	}
	for(int i=1;i<=m;i++)
	{
    
    
		int x;cin>>x;
		b.push_back(x);
	}
	int cnt=0;
	while(!a.empty() && !b.empty())
	{
    
    
		if(a.front()==b.front()) 
		{
    
    
			a.pop_front();
			b.pop_front();
		}
		else if(a.front()<b.front())
		{
    
    
			int a1=a.front();a.pop_front();
			int a2=a.front();a.pop_front();
			a.push_front(a1+a2);
			cnt++;
		}
		else if(a.front()>b.front())
		{
    
    
			int b1=b.front();b.pop_front();
			int b2=b.front();b.pop_front();
			b.push_front(b1+b2);
			cnt++;
		}
	}
	cout<<cnt;
	
    return 0;
}