CSU-ACM2017暑假集训2-二分搜索 hdu2141- Can you find it?

题目：

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 31 2 31 2 31 2 331410

Sample Output

Case 1:NOYESNO

           题目大意：问能否在A,B,C,三个数组内各选一个数，使得他们的和为给定的值X

           思路：先选两个数组比如A,B求其各个数相加之和，存到一个数组T中，再枚举数组C的数得到满足X-c=a+b的值，再在T数组中二分查找是否            存在X-c为A,B之和。复杂度O(n*n).

           代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int l,n,m;int a[555],b[555],c[555];int a_b[555*555];int a_b_norep[555*555];int num_ab;void init(){    int k,t;    k=t=0;    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]);    for(int i=0;i<l;i++)    {        for(int j=0;j<n;j++)        {            a_b[k++]=a[i]+b[j];        }    }    sort(a_b,a_b+k);    a_b_norep[t++]=a_b[0];    for(int i=1;i<k;i++)        if(a_b[i]!=a_b[i-1])            a_b_norep[t++]=a_b[i];    num_ab=t;}int bs(int x){    int l,r,mid;    l=0;    r=num_ab;    while(l<=r)    {        mid=(r-l)/2+l;        if(a_b_norep[mid]==x)            return 1;        else if(a_b_norep[mid]<x)            l=mid+1;        else            r=mid-1;    }    return 0;}int main(){    int t,x;    int no=0;    while(scanf("%d%d%d",&l,&n,&m)!=EOF)    {        init();        scanf("%d",&t);        printf("Case %d:\n",++no);        while(t--)        {            int flag=0;            scanf("%d",&x);            for(int i=0;i<m;i++)            {                if(bs(x-c[i]))                {                    flag=1;                    break;                }            }            if(flag)                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}