A. Rudolf and the Ticket

題解

簡單的二分應用，對於每個bi我們只需找到最大的ci使得bi+ci<=target即可

code

#include<bits/stdc++.h>
using namespace std;
int a[105],b[105];
int main(){
//    freopen("input.txt","r",stdin);
    int t;
    cin>>t;
    while (t--){
        int n,m,k,sum=0;
        cin>>n>>m>>k;
        for (int i=1;i<=n;i++) cin>>a[i];
        for (int i=1;i<=m;i++) cin>>b[i];
        sort(a+1,a+n+1);
        sort(b+1,b+m+1);
        for (int i=1;i<=n;i++){
            if (a[i]+b[1]>k) break;
            int l=1,r=m+1;
            while (l+1<r){
                int mid=l+(r-l>>1);
                if (a[i]+b[mid]<=k) l=mid;
                else r=mid;
            }
            sum+=l;
        }
        cout<<sum<<endl;
    }
    return 0;
}