Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 43435 Accepted Submission(s): 10508
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
题意：给A,B,C三个序列，分别从这三个序列中找三个数，若存在三个数之和等于X，则输出”YES”，否则输出”NO”
思路：三重循环的搜索肯定是不能用的。我的解法是将a+b的值存到ab数组中，将该数组排序。再用二分搜索判断x-c是否存在于该数组中，若存在，则证明存在x=a+b+c，输出’YES’即可。若对于所有c,都不存在x-c属于该数组,则证明不存在x=a+b+c，输出’NO’即可。
AC代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int sum[250010],a[505],b[505],c[505];
int div(int begin,int end,int ans) //二分查找函数
{
int left=begin,right=end;
while(left<=right)
{
int mid=left+right>>1;
if(sum[mid]==ans) return 1; //找到
if(sum[mid]>ans) right=mid-1;
else if(sum[mid]<ans) left=mid+1;
}
return 0;
}
int main(){
int k=1;
int i,j;
int l,n,m;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
printf("Case %d:\n",k);
for(i=1;i<=l;i++) scanf("%d",&a[i]);
for(i=1;i<=n;i++) scanf("%d",&b[i]);
for(i=1;i<=m;i++) scanf("%d",&c[i]);
int cnt=1;
for(i=1;i<=l;i++)
for(j=1;j<=n;j++)
sum[cnt++] = a[i]+b[j];
sort(sum+1,sum+cnt);
int t;
scanf("%d",&t);
while(t--)
{
int x,ans;
int flag=0;
scanf("%d",&x);
for(i=1;i<=m;i++)
{
ans=x-c[i];
if(div(1,cnt-1,ans))
{
flag=1;
break;
}
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
k++;
}
return 0;
}