最近感覺二分壓力超級大，所以和妹子一起做了二分的題目，好水的題目啊，可是沒辦法誰叫我們太弱了呢，繼續加油，我要變大牛，不要做菜鳥。
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
一看數據那麼大，就想到二分了，首先把二個變成一個再二分時間復雜度為nlog(n*n)這樣就不會超時了。二分算法就不說了，地球人都知道。
代碼：
[cpp]
<span style="font-family:FangSong_GB2312;font-size:18px;">#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int a[505],b[505],c[505];
int dp[250025];
int main()
{
int i,j,k,m,n,s,t,cas=0,p,flag;
while(scanf("%d%d%d",&k,&m,&n)!=EOF)
{
for(i=0;i<k;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]);
p=0;
for(i=0;i<k;i++)
for(j=0;j<m;j++)
dp[p++]=a[i]+b[j];
sort(dp,dp+p);
printf("Case %d:\n",++cas);
scanf("%d",&t);
while(t--)
{
scanf("%d",&s); flag=0;
for(i=0;i<n;i++)
{
int l=0,r=p-1,mid;
while(l<=r)
{
mid=(l+r)/2;
if(dp[mid]+c[i]==s){ flag=1;break;}
else if(dp[mid]+c[i]<s) l=mid+1;
else if(dp[mid]+c[i]>s) r=mid-1;
}
if(flag) break;
}
if(flag) printf("YES\n");
else printf("NO\n");
}
}
return 0;
}
</span>
