啊希吧！第一场队内赛总结

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<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 31 2 31 2 31 2 331410

Sample Output

Case 1:NOYESNO

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 1001int 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;}

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