HDU 2141 Can you find it?(二分）

<strong>                                                 <span style="font-size:32px;">Can you find it?</span></strong><span style="font-size:24px;"><strong>Description</strong></span><span style="font-family:SimSun;font-size:18px;">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. </span><span style="font-size:24px;"><strong>Input</strong></span><span style="font-family:SimSun;font-size:18px;">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. </span><strong><span style="font-size:24px;">Output</span></strong><span style="font-family:SimSun;font-size:18px;">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". </span><span style="font-size:24px;"><strong>Sample Input</strong></span><span style="font-size:18px;">3 3 31 2 31 2 31 2 331410 </span> <strong><span style="font-size:24px;">Sample Output</span></strong><span style="font-size:18px;">Case 1:NOYESNO </span>

解决代码：

<span style="font-size:14px;">#include<stdio.h>#include<algorithm>using namespace std;#define K 505int LN[K*K];int Find(int LN[],int h,int t){    int left,right,mid;    left=0;    right=h-1;    while(left<=right)    {        mid=(left+right)/2;        if(LN[mid]==t)          return 1;        else if(LN[mid]>t)          right=mid-1;        else if(LN[mid]<t)          left=mid+1;    }    return 0;}int main(){    int i,j,count=1,x;    int L[K],N[K],M[K],S,n,m,l;    while(scanf("%d%d%d",&l,&n,&m)!=EOF)    {        int h=0;        for(i=0;i<l;i++)         scanf("%d",&L[i]);        for(i=0;i<n;i++)          scanf("%d",&N[i]);        for(i=0;i<m;i++)          scanf("%d",&M[i]);        for(i=0;i<l;i++)          for(j=0;j<n;j++)           LN[h++]=L[i]+N[j];        sort(LN,LN+h);        scanf("%d",&S);        printf("Case %d:\n",count++);        for(i=0;i<S;i++)        {            scanf("%d",&x);            int p=0;            for(j=0;j<m;j++)            {                int a=x-M[j];                 if(Find(LN,h,a))                {                    printf("YES\n");                    p=1;                    break;                }            }            if(!p)              printf("NO\n");        }    }    return 0;}</span>