A - Can you find it?解题报告

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A - Can you find it?

Time Limit:3000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status

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

这个题目的意思呢就是从三组数中分别拉一个数出来，如果和==X的话就输出“YES”，否则输出“NO”；一开始虽然知道用硬的来时会超时，可是还是抱有侥幸的心态，结果TLE；还是用二分吧。这个二分首先将两个数组的和加起来，就是先开一个250001的数组，然后sort排序，然后呢就查找咯；
看代码把：

#include<iostream>#include<algorithm>using namespace std;int main(){int a[501],b[501],c[501],d[250001],L,N,M,k=0,i,j,qq=1,n,S;while(cin>>L>>N>>M){k=0;for(i=0;i<L;i++)cin>>a[i];for(i=0;i<N;i++)cin>>b[i];for(i=0;i<M;i++)cin>>c[i];for(i=0;i<L;i++)for(j=0;j<N;j++)d[k++]=a[i]+b[j];sort(d,d+k);cin>>n;cout<<"Case "<<qq<<':'<<endl;for(i=0;i<n;i++){cin>>S;for(j=0;j<M;j++){int temp=S-c[j];int left=0,right=k-1,mid;while(left<right){mid=(left+right)/2;if(d[mid]==temp){cout<<"YES"<<endl;goto end;}else if(temp>d[mid]){left=mid+1;}else{right=mid;}}}cout<<"NO"<<endl;end:;}qq++;}return 0;}