HDOJ 2141 Can you find it?

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 

 

思路：把前两个数组加起来组成一个数组，并排序，然后对其下标进行二分，d[mid]与x-c[i]比较，只要存在一个就break;

代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;long long a[510],b[510],c[510];long long d[510*510];int main(){int L,M,N,s;int k=1;while(scanf("%d%d%d",&L,&N,&M)!=EOF){for(int i=1;i<=L;i++){scanf("%lld",&a[i]);}for(int i=1;i<=N;i++){scanf("%lld", &b[i]);}for(int i=1;i<=M;i++){scanf("%lld",&c[i]);}int t=0;for(int i=1;i<=L;i++){for(int j=1;j<=N;j++){d[++t]=a[i]+b[j];}}sort(d+1,d+t+1);scanf("%d",&s);printf("Case %d:\n",k++);while(s--){    bool flag=false;long long  x;scanf("%lld",&x);for(int i=1;i<=M;i++){int l=1,r=t;//对下标二分，每次都找中间的那个下标，比较d[mid]与目标； while(l<=r){int mid=(l+r)/2;if(d[mid]==x-c[i]){flag=true;break;}else if(d[mid]<x-c[i]){l=mid+1;}else{r=mid-1;}}if(flag)//只要找到一个就可以终止；    break;}if(flag)   printf("YES\n");else    printf("NO\n");}}return 0;}