HDU 2141 Can you find it? 二分查找

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 21485    Accepted Submission(s): 5446

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

 

Recommend

威士忌

 

题目大意：输入三个数l, m, n。代表序列A(a1,a2,a3....ai)，B(b1,b2,b3...bi)，C(c1,c2,c3...ci)，分别有l,m,n个数，接下来输入3个序列，再输入S，代表S组数据，接下来输入S组数据，判断这些数能否等于AI+BJ+CK(i,j,k任意)

 

思路：开始用暴力3个for，会超时。想到用二分，可以用一个新数组sum存入a+b所有可能，用sort排序，再判断x-c能否在sum数组找到。

 

#include<iostream>
#include<algorithm>
int a[510], b[510], c[510],sum[250010];
using namespace std;
int f(int x,int z,int y)//二分注意加一 
{
    int mid; 
    while(z<=y)
    {
        mid=(z+y)/2;
        if(sum[mid]==x) return 1;
        else if(sum[mid]>x)
        y=mid-1;
        else if(sum[mid]<x)
        z=mid+1;
    }
    return 0;
}
int main()
{
    int l,m,n,z,i,j,t,k=1;
    while(scanf("%d%d%d", &l, &m, &n)!=EOF)
    {
        for(i=0;i<l;i++)scanf("%d",&a[i]);//一开始l,m,n写乱了,WA了好几遍，玛德智障。。。
        for(i=0;i<m;i++)scanf("%d",&b[i]);
        for(i=0;i<n;i++)scanf("%d",&c[i]);
        for(i=0;i<l;i++)
            for(j=0;j<m;j++)
                sum[i*m+j]=a[i]+b[j];
        sort(sum,sum+l*m);
        printf("Case %d:
",k++);
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&z);
            for(j=0;j<n;j++)
                if(f(z-c[j],0,l*m))
                break;
            if(j!=n)
            printf("YES
");
            else
            printf("NO
");
        }
    }
    return 0;
}