HDU 2141 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 3
1 2 3
1 2 3
1 2 3
3
1
4
10
Sample Output
Case 1:
NO
YES
NO

题意：

有三个数组A、B、C，问是否存在Ai、Bi、Ci，使得Ai+Bi+Ci=X。

分析：

典型的二分题。如果用三个for循环暴力解决，时间复杂度为O(n^3)。考虑对其中两个数组求和并排序，再枚举将另一个数组，对和数组进行二分查找A+B=X-C。

#include <iostream>#include <algorithm>using namespace std;const int max1=505;int l,n,m,s,x,bci;   //bci即B、C数组合并后的元素数量int a[max1],b[max1],c[max1*max1];bool findx(){    for(int i=0;i<l;++i)    {        int lo=0,hi=bci-1,mid;        while(lo<=hi)        {            mid=(lo+hi)>>1;            if(c[mid]+a[i]==x) return true;            else if(c[mid]+a[i]>x)  hi=mid-1;            else lo=mid+1;        }    }    return false;}int main(){    int d=0;    while(cin>>l>>n>>m)    {        for(int i=0;i<l;++i)    cin>>a[i];        for(int i=0;i<n;++i)    cin>>b[i];        bci=0;        int temp;        for(int i=0;i<m;++i){            cin>>temp;            for(int j=0;j<n;++j){                c[bci++]=b[j]+temp;            }        }    //  c数组存放b+c        sort(c,c+bci);        cin>>s;     cout<<"Case "<<++d<<":"<<endl;        while(s--)        {            cin>>x;            //枚举a数组，二分c数组            if(findx())    cout<<"YES"<<endl;            else    cout<<"NO"<<endl;        }    }    return 0;}

下面利用binary_search()，更简洁，但用时稍多。

#include <iostream>#include <algorithm>using namespace std;const int max1=505;int a[max1],b[max1],c[max1],bc[max1*max1];int main(){    int d=0;    int l,n,m,s,x,bci=0;    while(cin>>l>>n>>m)    {        bci=0;        for(int i=0;i<l;++i)    cin>>a[i];        for(int i=0;i<n;++i)    cin>>b[i];        for(int i=0;i<m;++i)    cin>>c[i];        for(int i=0;i<n;++i)            for(int j=0;j<m;++j)                bc[bci++]=b[i]+c[j];        sort(bc,bc+bci);        cin>>s;     cout<<"Case "<<++d<<":"<<endl;        while(s--)        {            cin>>x;            int cnt=0;            for(int i=0;i<l;++i)            {                if(binary_search(bc,bc+bci,x-a[i]))                    {cnt=1;break;}            }            if(cnt)    cout<<"YES"<<endl;            else    cout<<"NO"<<endl;        }    }    return 0;}