HDU 2141 二分查找

还有http://blog.csdn.net/libin56842/article/details/17336981没做

数据结构http://blog.csdn.net/olga_jing/article/details/50912239

字典树

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

 

题目意思是：给你3个数组a[],b[],c[],再给你s个询问，对于每个询问X，若能在a,b,c三个数组各找1个数字，使得他们的和为X，则输出YES否则输出NO。我的思路是在a，b数组中按顺序各取一个数加起来放入新数组num中，对num数组和c数组进行排序。排序后，遍历排序后的c数组，查询num数组中是否有X-ci，若有则输出YES，否则输出NO

最重要的是用对数据类型，保存a与b之和要用long long或__int64

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<iostream>
#include<math.h>
#define N 505
using namespace std;
__int64 num[N*N];
int mybinary_search(int len,int target)
{
    int L,R,mid=0;
    L=0;
    R=len-1;
    while(L<=R)
    {
        mid=(L+R)>>1;
        if(num[mid]>target)R=mid-1;
        else if(num[mid]<target)L=mid+1;
        else return mid;
    }
    return -1;
}

int main()
{
    int i,j,k,l,n,m,query,cnt1,cnt2;
    int a,b,c,flag;
    int  aa[505],bb[505],cc[505];
    cnt2=0;
    while(scanf("%d%d%d",&l,&m,&n)!=EOF)
    {
        cnt1=0;
        for(i=0; i<l; i++)scanf("%d",&aa[i]);
        for(i=0; i<m; i++)scanf("%d",&bb[i]);
        for(i=0; i<n; i++)scanf("%d",&cc[i]);
        for(i=0; i<l; i++)
            for(j=0; j<m; j++)
                num[cnt1++]=aa[i]+bb[j];
        sort(num,num+cnt1);
        sort(cc,cc+n);
        scanf("%d",&k);
        printf("Case %d:
",++cnt2);
        while(k--)
        {
            scanf("%d",&query);
            flag=0;
            if(query<num[0]+cc[0]||query>num[cnt1-1]+cc[n-1])
            {
                printf("NO
");
                continue;
            }
            for(j=0; j<n; j++)
            {
                if(mybinary_search(cnt1,query-cc[j])!=-1)
                {
                    printf("YES
");
                    flag=1;
                    break;
                }
            }
            if(!flag)printf("NO
");
        }
    }

    return 0;
}