二分 O(n * logm) 145ms
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m, x;
int a[N], b[N];
int main() {
scanf("%d%d%d", &n, &m, &x);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
for (int i = 0; i < m; i++) scanf("%d", &b[i]);
for (int i = 0; i < n; i++) {
int val = x - a[i];
int l = 0, r = m - 1;
while (l < r) {
int mid = l + r >> 1;
if (b[mid] >= val) r = mid;
else l = mid + 1;
}
if (b[l] == x - a[i]) {
printf("%d %d", i, l);
break;
}
}
return 0;
}
双指针 O(n + m) 123ms
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m, x;
int a[N], b[N];
int main() {
scanf("%d%d%d", &n, &m, &x);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
for (int i = 0; i < m; i++) scanf("%d", &b[i]);
for (int i = 0, j = m - 1; i < n && j >= 0; i++) {
while (j >= 0 && b[j] + a[i] > x) j--;
if (a[i] == x - b[j]) {
printf("%d %d", i, j);
break;
}
}
return 0;
}