原題：
D. Vasya and Arrays
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1,10,100,1000,10000] Vasya can obtain array [1,1110,10000], and from array [1,2,3] Vasya can obtain array
[6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i Ai=Bi.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B
become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and Bequal.
Input
The first line contains a single integer n (1≤n≤3⋅10^5) — the length of the first array.
The second line contains n integers a1,a2,⋯,an(1≤ai≤10^9) — elements of the array A.
The third line contains a single integer m (1≤m≤3⋅10^5) — the length of the second array.
The fourth line contains m integers b1,b2,⋯,bm (1≤bi≤10^9) - elements of the array B.
Output
Print a single integer — the maximum length of the resulting arrays after some operations were performed on arrays
A and B in such a way that they became equal.
If there is no way to make array equal, print “-1”.
Examples
input
5
11 2 3 5 7
4
11 7 3 7
output
3
input
2
1 2
1
100
output
-1
input
3
1 2 3
3
1 2 3
output
3
中文：
給你兩個序列，給你一個操作，問你最少多少步操作可以使這兩個序列相同，兩個序列相同就是這兩個序列長度相同而且對應下標的數字相同。
操作是可以將相鄰的兩個數字合併，例如[1,2,3,4,5]，可以將2和3合併爲5，結果是[1,5,4,5]。
如果不能通過操作使得兩個序列相同，輸出-1.
代碼：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=300005;
int n,m;
ll a[maxn],b[maxn],sa[maxn],sb[maxn];
int main()
{
ios::sync_with_stdio(false);
while(cin>>n)
{
memset(sa,0,sizeof(sa));
memset(sb,0,sizeof(sb));
for(int i=1;i<=n;i++)
{
cin>>a[i];
sa[i]=sa[i-1]+a[i];
}
cin>>m;
for(int i=1;i<=m;i++)
{
cin>>b[i];
sb[i]=sb[i-1]+b[i];
}
if(sa[n]!=sb[m])
{
cout<<-1<<endl;
continue;
}
if(n==1||m==1)
{
if(sa[n]==sb[m])
cout<<1<<endl;
else
cout<<-1<<endl;
continue;
}
int i=1,j=1,flag=0,ans=n;
while(true)
{
if(i==n&&j==m)
break;
if(a[i]==b[j])
{
if(i<n)
i++;
if(j<m)
j++;
}
else
{
int pa=i,pb=j;
while(sa[i]-sa[pa-1]!=sb[j]-sb[pb-1])
{
if(sa[i]-sa[pa-1]>sb[j]-sb[pb-1])
{
if(j==m)
{
flag=-1;
break;
}
j++;
}
if(sa[i]-sa[pa-1]<sb[j]-sb[pb-1])
{
if(i==n)
{
flag=-1;
break;
}
i++;
}
}
ans-=(i-pa);
if(i<n)
i++;
if(j<m)
j++;
}
if(flag==-1)
break;
}
if(flag==-1)
cout<<-1<<endl;
else
cout<<ans<<endl;
}
return 0;
}
思路：
首先可以判斷的是，如果兩個序列A和B的總和不相同，那麼肯定不能通過操作使得兩個序列相等。
如果兩個序列的總和相同，可以設置兩個下標i和j，分別指向序列A和序列B的開頭，依次判斷下標對應值是否相等，如果相等，下標前移。否則，指向的元素值小的那個下標向前累加，如果累加和超過另一個下標所指向的元素，則另外一個下標也向前累加，直到兩個累加和相同。如果兩個下標移動到最後也沒使兩個累加和相同，那麼輸出-1，否則計數累加過的元素個數再減去1就是答案。