啊希吧！第一场队内赛总结

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">今天学长讲完二分法打了一场队内赛，</span>

6道题目截止到刚才AC了5道，

还有一道真的不会 = =

讲一下总结

其实rank可以更高

B题换了个double就过了

之前无限WA。。。

然后，A题二分写cuo了= =

然后强烈的感觉就是自己真的是弱渣，

可以随便被虐的那种 = =

时间复杂度这个东西知道有一年多了

然而真的有直观的感受还是今天= =

于是我就拿A题出来说一下子。

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 

三个序列开的大小分别是 500 500 500 和序列开的是1000

相乘的话 就是125*10^9

1s大概是 10^8的样子

暴力肯定TLE

于是想到一种方法

A B所有的相加情况变成一个序列E

然后和序列D减去另一个序列C

再在E序列中去查找

这样的话

时间复杂度就变成了

log(500*500)*500*1000

直接降了好几个数量级

嗯 然后二分查找的模板还是要记清楚的

比如是low>=high

最后贴代码

#include <stdio.h>
#include <algorithm>
using namespace std;
#define maxn 1001

int a[maxn],b[maxn],c[maxn],x[maxn],d[maxn*maxn];

int icq(int a[],int n,int key)
{
    int low=0;
    int high=n-1;
    int mid;
    while(low<=high)
    {
        mid=(low+high)/2;
        if(a[mid]<key)
        {
            low=mid+1;
        }
        else if(a[mid]>key)
        {
            high=mid-1;
        }
        else if(a[mid]==key)
            return 1;
    }
    return -1;

}


int main()
{
    int d0,m,n,s,i,j,k,l,flag;
    d0=0;
    while(scanf("%d %d %d",&l,&m,&n)!=EOF)
    {
        d0++;
        for(i=0; i<l; i++)
            scanf("%d",&a[i]);
        for(i=0; i<m; i++)
            scanf("%d",&b[i]);
        for(i=0; i<n; i++)
            scanf("%d",&c[i]);
            scanf("%d",&s);
        for(i=0; i<s; i++)
            scanf("%d",&x[i]);
        printf("Case %d:\n",d0);

        k=0;
        for(i=0; i<l; i++)
            for(j=0; j<m; j++)
            {
                d[k]=a[i]+b[j];
                ++k;
            }
        sort(d,d+k);


        for(j=0; j<s; j++)
        {
            flag=0;
            i=0;
            for(i=0; i<n; i++)
            {
                l=x[j]-c[i];
                if(icq(d,k,l)==1)
                {
                    printf("YES\n");
                    flag=1;
                    break;
                }
            }
            if(flag==0)printf("NO\n");
        }

    }
    return 0;
}