看到大家都队列模拟，但前缀和也可以吧。代码如下： ```cpp #include <bits/stdc++.h> using namespace std; const int N = 1e5 + 10; typedef long long ll; ll a[N], b[N]; int main() { int n, m; cin >> n >> m; for (int i = 1; i <= n; i++) { ll tmp; cin >> tmp; a[i] = a[i - 1] + tmp; } for (int i = 1; i <= m; i++) { ll tmp; cin >> tmp; b[i] = b[i - 1] + tmp; } ll res = 0; int index1 = 1, index2 = 1; int pre_index1 = 0, pre_index2 = 0; while (index1 <= n && index2 <= m) { if (a[index1] - a[pre_index1] == b[index2] - b[pre_index2]) { res += index1 + index2 - pre_index1 - pre_index2 - 2; pre_index1 = index1; pre_index2 = index2; index2++; index1++; } else if (a[index1] - a[pre_index1] < b[index2] - b[pre_index2]) { index1++; } else { index2++; } } cout << res; return 0; } ```

emmmm既然要最后两个数组一样 那么他们的长度必须一样，而且只能有合并操作，那么合并一次长度减一，在第二个数组不进行合并的情况下，直接用两个数组长度相减就是答案。而且由题目可以得到，第二个数组不进行任何操作的情况也是正解 ``` #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef struct point{ int x; int y; }point; int main(){ int n,m; cin>>n>>m; int arr[n+10]; int brr[m+10]; for(int i=1;i<=n;i++) cin>>arr[i]; for(int j=1;j<=m;j++) cin>>brr[j]; cout<<abs(n-m)<<endl; return 0; } ```

直接模拟即可, 代码很简单, 你一定能看明白的 ```cpp #include <cstdio> #include <list> using namespace std; int main() { int res = 0; int n = 0, m = 0; scanf("%d %d", &n, &m); list<int> arr; list<int> brr; for (int i = 0, j; i < n; ++i) { scanf("%d", &j); arr.push_back(j); } for (int i = 0, j; i < m; ++i) { scanf("%d", &j); brr.push_back(j); } while (arr.size() && brr.size()) { if (*arr.rbegin() == *brr.rbegin()) { arr.pop_back(); brr.pop_back(); } else { ++res; if (*arr.rbegin() > *brr.rbegin()) { // 合并 brr 的 int x = *brr.rbegin(); brr.pop_back(); int& y = *brr.rbegin(); y += x; } else { // 合并 arr 的 int x = *arr.rbegin(); arr.pop_back(); int& y = *arr.rbegin(); y += x; } } } printf("%d\n", res); return 0; } ```

``` #include <bits/stdc++.h> using namespace std; queue<int> a, b; signed main(){ int n, m; cin >> n >> m; for(int i = 1; i <= n; ++i){ int x; cin >> x; a.push(x); } for(int i = 1; i <= m; ++i){ int x; cin >> x; b.push(x); } int cnt = 0; while(a.size() || b.size()){ int top1 = a.front(), top2 = b.front(); a.pop(), b.pop(); while(top1 != top2){ if(top1 < top2){ top1 += a.front(); ++cnt; a.pop(); }else{ top2 += b.front(); ++cnt; b.pop(); } } } cout << cnt << endl; return 0; } ```

include define int long long using namespace std; int n,m,ans=0; int a[100001],b[100001]; ``` ` ``` ```` signed main() { cin >>n>>m; for(int i=0;i b[j])//同理 { j++; ans++; if(j<m) b[j]+=b[j-1]; } } cout <<ans<<endl; return 0; }