HDU

Can you find it?

Problem Description
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

题目大意：在A、B、C三个数组中各选出一个数，看能否使得三个数之和为给定的数。只需讲其中两个数组中任意两个元素相加得到一个新的数组，再与剩下的数组中每一个元素进行二分即可

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 510;int a[maxn], b[maxn], c[maxn], d[maxn*maxn];int cnt;int bs(int x){    int hi;    int lo;    int mid;    lo = 0;    hi = cnt-1;    while(hi>=lo)    {        mid = (lo+hi) >> 1;        if(d[mid]<x)            lo = mid + 1;        else if(d[mid]>x)            hi = mid - 1;        else            return 1;    }    return 0;}int main(){    int l, m, n, k, s;    int flag = 1;    while(cin >> l >> m >> n)    {        for(int i = 0; i < l; i++)        {            cin >> a[i];        }        for(int i = 0; i < m; i++)        {            cin >> b[i];        }        for(int i = 0; i < n; i++)        {            cin >> c[i];        }        cnt = 0;        for(int i = 0; i < m; i++)        {            for(int j = 0; j < n; j++)            {                d[cnt++] = b[i] + c[j];            }        }        sort(d, d+cnt);        sort(a, a+l);        cin >> s;        printf("Case %d:\n", flag++);        int sum;        while(s--)        {            cin >> sum;            int ok = 0;            /*if(a[0]+d[0]>sum || a[l-1]+d[cnt-1]<sum)            {                cout << "NO" << endl;                continue;            }*/            //这段优化代码加进去就错了。。我也不知道为什么，有人如果能看出来原因请告诉我            for(int j = 0; j < l; j++)            {                if(bs(sum-a[j]))                {                    ok = 1;                    break;                }            }            if(ok)                cout << "YES" << endl;            else                cout << "NO" << endl;        }    }    return 0;}