C++ 代码

//对每一个r，都存在l恰好使和小于等于x，随着r增大，l减小。所以递增r时，只需递减l即可，无需回退
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N], b[N];
int main()
{
    int n, m, x;
    cin >> n >> m >> x;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < m; i++) cin >> b[i];
    int l = n - 1, r = 0;
    for (; r < m; r++)
    {
        while (a[l] + b[r] > x) l--;
        if (a[l] + b[r] == x) break;
    }
    cout << l << ' ' << r;
    return 0;
}