题目链接：
Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 38634 Accepted Submission(s): 9395
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
题目大意：
多组数据输入
第一行是三个序列各自元素的个数
第2，3，4是三个序列
第4行是x的个数
下面的x行是多种情况的x值
分析：
三个序列三个数组，把前面两个数组组合为一个数组
比如1 2 3，1 2 3，这两个序列组合为一个数组‘
1+1，1+2，1+3，2+1，2+2，2+3，3+1，3+2，3+3，
计算一下就是
2，3，4，3，4，5，4，5，6，有重复的但是不用去掉，称这个序列为t
然后与第三个序列组合
1 ，2，3，（第三个序列）
然后将t序列升序排序，t与第三个序列组合的时候，t序列采用二分寻找数据
注意点：
1.不能采用set去重，不然会超内存（第一次遇到超内存，好激动啊啊啊啊啊）
2.对t序列不能采用for循环寻找数据，要采用排好序之后再二分查找数据，不然会超时
代码如下：
#include<bits/stdc++.h>
using namespace std;
bool cmp(int a,int b)
{
return a<b;
}
int main()
{
int n1,n2,n3,n=;
while(~scanf("%d %d %d",&n1,&n2,&n3))
{
int a[n1],b[n2],c[n3];
for(int i=; i<n1; i++)
{
scanf("%d",&a[i]);
}
for(int j=; j<n2; j++)
{
scanf("%d",&b[j]);
}
for(int k=; k<n3; k++)
{
scanf("%d",&c[k]);
}
int xn;
int t[n1*n2];
scanf("%d",&xn);
int x[xn];
for(int i=; i<xn; i++)
{
scanf("%d",&x[i]);
}
int y=;
for(int i=; i<n1; i++)
{
for(int j=; j<n2; j++)
{
int z=a[i]+b[j];
t[y]=z;
y++;
}
}
sort(t,t+(n1*n2),cmp);
printf("Case %d:\n",n);
for(int i=; i<xn; i++)
{
int f=;
for(int k=;k<n3;k++)
{
int l=,h=n1*n2-;
while(l<=h)
{
int mid=(l+h)/; if(t[mid]==x[i]-c[k])
{
f=;
break;
}else if(t[mid]>x[i]-c[k])
{
h=mid-;
}else if(t[mid]<x[i]-c[k])
{
l=mid+;
}
}
if(f==)
break;
}
if(f==)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
n++;
}
return ;
}
哈哈哈哈哈哈哈哈哈哈
今天真的是超级开心
不准吐槽。。。。。
。。。。。。。。。。