HDU 2141 Can you find it?（二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 35836    Accepted Submission(s): 8845

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

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

 

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题意：给你三个数组，再给你一些数，判断三个数组内是否有Ai+Bi+Ci==x.如果有输出YES,否则输出NO、

思路：先将前两个数组两两相加构成一个新数组，快排。然后新数组和第三个数组相加，二分，不断用中间的和第三个数组的元素相加，直到得x或者跳出循环为止。

 

#include<cstdio>  
#include<iostream>  
#include<algorithm>  
#include<cmath>  
#define xps 1e-12  
using namespace std;
long long a[510], b[510], c[510], d[250010];
int main()
{
    long long l, m, n, o = 1;
    while (cin >> l >> m >> n)
    {
        for (int i = 0; i<l; i++)//输入三个数组   
            cin >> a[i];
        for (int i = 0; i<m; i++)
            cin >> b[i];
        for (int i = 0; i<n; i++)
            cin >> c[i];
        int k = 0;
        for (int i = 0; i<l; i++)
            for (int j = 0; j<m; j++)
                d[k++] = a[i] + b[j];//前两个数组两两相加得到新数组   
        sort(d, d + k);//从小到大排序   
        int s;
        cin >> s;
        cout << "Case " << o++ << ":" << endl;//输出第几个测试实例   
        while (s--)
        {
            long long x;
            cin >> x;
            bool sign = false;
            for (int i = 0; i<n; i++)//第三个数组和新数组mid相加判断   
            {
                int l = 0, r = k - 1;
                while (l <= r)
                {
                    int mid = (l + r) / 2;
                    if (d[mid] + c[i] == x)
                    {
                        sign = 1;
                        break;
                    }
                    else if (d[mid] + c[i] < x)
                    {
                        l = mid+1;
                    }
                    else r = mid-1;
                }
                if (sign) break;
            }
            if (sign)
                cout << "YES" << endl;
            else
                cout << "NO" << endl;
        }
    }
    return 0;
}