【最大通过数——二分】

题目

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int N = 2e5 + 10;

int n, m, k;
ll a[N], b[N];
int idx1, idx2;
ll ans;

struct node
{
    ll val, cnt;
    bool operator<(const node &t) const
    {
        return val < t.val;
    }
} cnta[N], cntb[N];
int main()
{
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", a + i);
        a[i] += a[i - 1];
        if (a[i] <= 1e9)
            cnta[++idx1] = {a[i], i};
    }
    for (int i = 1; i <= m; i++)
    {
        scanf("%d", b + i);
        b[i] += b[i - 1];
        if (b[i] <= 1e9)
            cntb[++idx2] = {b[i], i};
    }

    for (int i = 1; i <= idx1; i++)
    {
        if (cnta[i].val > k)
            break;
        int left = k - cnta[i].val;
        int idx = upper_bound(cntb + 1, cntb + idx2 + 1, (node){left, 0}) - cntb - 1;
        ans = max(ans, cnta[i].cnt + cntb[idx].cnt);
    }
    int idx = upper_bound(cntb + 1, cntb + idx2 + 1, (node){k, 0}) - cntb - 1;
    ans = max(ans, cntb[idx].cnt);
    
    printf("%lld", ans);
    return 0;
}