Can you find it? 【二分】

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.
Input
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.
Output
For each case, firstly you have to print the case number as the form “Case d:”, then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print “YES”, otherwise print “NO”.
Sample Input
3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10
Sample Output
Case 1:
NO
YES
NO

记得 白书上好像有和这个类似的。 就是要先将式子变下形，
A+B = X-C ；
并且让A+B 合在一个数组里； 【不是很难，不过一定要理解这个二分枚举】
代码

#include<bits/stdc++.h>#define LL long longusing namespace std;const int MAXN = 500+10;const int MAXM = 3000000;const int inf = 0x3f3f3f3f;const double pi= acos(-1.0);const double eps = 1e-5;/*------------------------------*/int l,n,m;int a[MAXN],b[MAXN],c[MAXN];int d[MAXM];int main(){    int kk=1;    while(~scanf("%d%d%d",&l,&n,&m)){        for(int i=0;i<l;i++)            scanf("%d",&a[i]);        for(int i=0;i<n;i++)            scanf("%d",&b[i]);        for(int i=0;i<m;i++)            scanf("%d",&c[i]);             printf("Case %d:\n",kk++);            int s=0;        for(int i=0;i<l;i++){            for(int j=0;j<n;j++){                 d[s++]=a[i]+b[j];            }        }        int ri=s;        sort(d,d+s);        int q;cin>>q;        while(q--){        int a;scanf("%d",&a);            int i;   int mid;            for(i=0;i<m;i++){                int o=a-c[i];                int L=0;int R=ri-1;                int ss;                while(L<=R){                    mid=(L+R)>>1;                    if(d[mid]>=o) { ss=d[mid]; R=mid-1;}                    else L=mid+1;                }                //printf("ss=== %d \n",ss);                if(ss==o) break;            }            if(i==m) puts("NO");            else puts("YES");        }    }    return 0;}