You can see that the more books you read on the desk A, the less books you will read on the desk B. So there's no need to use binary search, you can simply calculate the number of books on the dest B from the previous value.

The time complexity: $$$O(n + m)$$$, a bit faster the binary search $$$O(n\log{m})$$$.

#include <cstdio>
#include <vector>
#include <algorithm>

int main()
{
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    std::vector<int> a(n + 1), b(m + 1);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
    int ans = 0;
    long long suma = 0, sumb = 0;
    for (int i = 1; i <= m; i++) sumb += b[i];
    for (int i = 0, j = m; i <= n; i++)
    {
        suma += a[i];
        while (j > 0 && suma + sumb > k)
        {
            sumb -= b[j];
            j--;
        }
        if (suma + sumb > k) break;
        ans = std::max(ans, i + j);
    }
    printf("%d\n", ans);
}

Thank you !