HDU 2141 Can you find it? (二分法)

Can you find it?

Time Limit:3000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

 

 

Description

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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.

 

 

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Input

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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.

 

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Output

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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".

 

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Sample Input

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3 3 3

1 2 3

1 2 3

1 2 3

3

1

4

10

 

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Sample Output

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Case 1:

NO

YES

NO

 

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Analysis

　　像这种直接做找不到算法的题，一般都是用枚举、搜索这种“暴力”算法。

　　 这一题就是把 A+B 所有情况都存下来，对每一个 X ，把 C 移过来，令 X2 = X - C ，在 A+B 的数组中二分查找 X2 的值，如果找到一样的，输出YES，全部找不到，输出NO。

 

TIPs

• 在二分查找之前，一定要对 A+B 数组进行排序
• 二分的退出条件：可以用for循环循环100次，可以达到 2¹⁰⁰。如果那时找不到，则一定没有解了。

　 　　　　　　　　也可以在循环中：

　　　　　　　　　　while( front <= back )

　　　　　　　　　　{

　　　　　　　　　　　　　　　　 ......

　　　　　　　　　　　if( F[mid] > Z ) back = mid - 1;

　　　　　　　　　　　if(F[mid] < Z ) front = mid + 1;

　　　　　　　　　　　......

　　　　　　　　　　　......

　　　　　　　　　　}

 

 

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

using namespace std;

int ab[500*500+1];
int a[501],b[501],c[501];

int main()
{
    int na,nb,nc,nx,i,j,kkk=0;
    int x;

    while(~scanf("%d%d%d",&na,&nb,&nc))
    {

        for(i=0;i<na;i++)
            scanf("%d",&a[i]);
        for(i=0;i<nb;i++)
            scanf("%d",&b[i]);
        for(i=0;i<nc;i++)
            scanf("%d",&c[i]);

        int k=0,s,m,t;
        for(i=0;i<na;i++)
            for(j=0;j<nb;j++)
                ab[k++] = a[i] + b[j];
        sort(ab,ab+na*nb);

        scanf("%d",&nx);
        int flag = 0;
        printf("Case %d:\n",++kkk);
        for(i=0;i<nx;i++)
        {
            scanf("%d",&x);
            for(j=0;j<nc;j++)
            {
                int x2;
                x2 = x - c[j];
                flag = 0;
                s = 0;
                t = na * nb - 1;
                m = (s + t)/2;
                if(ab[m] == x2) flag = 1;
                while(s <= t)
                {
                    if(ab[m] > x2)
                        t = m-1;
                    else s = m+1;
                    m = (s + t)/2;
                    //printf("s = %d t = %d\n",s,t);
                    if(ab[m] == x2)
                    {
                        flag = 1;
                        break;
                    }
                }
                if(flag)
                {
                    printf("YES\n");
                    break;
                }
                else continue;
            }
            if(!flag) printf("NO\n");
        }
    }
    return 0;
}