只过了#11，样例已过，玄关

_mei_tou_nao_ @ 2024-11-08 15:51:25

#include <bits/stdc++.h>
using namespace std;
const int N = 1e8 + 5;
long long n, m, a[N], b[N], mid, ans;

int main() {
    cin >> m >> n;
    for (int i = 1; i <= m; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
        cin >> b[i];
    }
    sort(a + 1, a + m + 1);
    for (int j = 1; j <= n; j++) {
        long long l = 0, r = n + 1;
        while (l < r) {
            mid = (l + r) / 2;
            if (a[mid] <= b[j]) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        ans += min(abs(a[l] - b[j]), abs(a[l - 1] - b[j]));
    }
    cout << ans;
    return 0;
}

by dqh_0616 @ 2024-11-08 16:20:24

输入数据太大，循环会TLE，建议用二分。

by dqh_0616 @ 2024-11-08 16:37:42

#include<bits/stdc++.h>
using namespace std;
int n,m,a[100000],b[100000];
long long ans;
int main(){
    scanf("%d %d",&n,&m);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    for(int i=0;i<m;i++)
        scanf("%d",&b[i]);
    sort(a,a+n);
    for(int i=0;i<m;i++){
        int fs=lower_bound(a,a+n,b[i])-a;
        if(fs==n)
            ans+=b[i]-a[n-1];
        else if(fs==0)
            ans+=a[0]-b[i];
        else
            ans+=min(abs(a[fs]-b[i]),abs(b[i]-a[fs-1]));
    }
    printf("%ld",ans);
    return 0;
}

by _mei_tou_nao_ @ 2024-11-08 23:15:38

@dqh0616 其实我用循环打了个二分，下次应该说清楚不是TLE的