hdoj2141Can you find it?

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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 31 2 31 2 31 2 331410

Sample Output

Case 1:NOYESNO

题意：有三个集合，给一个n，分别从三个集合中抽取一个数，使这三个数的和等于n，问能不能找到这样的三个数.

思路：将这前两个集合（因为害怕超时，所以先忽略第三个集合）中分别抽取的两个数的和的所有情况按从小到大的顺序放入一个数组中，记录长度为k，然后使左边界为0，右边界为k，然后用二分搜索.

代码如下：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;bool cmp(int x,int y){return x<y;}int st1[550],st2[550],st3[550];int sum[550*550],k;bool judge(int x){int l,r,mid;l=1,r=k;while(l<=r){mid=(l+r)>>1;if(sum[mid]>x)r=mid-1;else if(sum[mid]<x)l=mid+1;elsereturn true;}return false;}int main(){int a,b,c,x,t=0;while(scanf("%d%d%d",&a,&b,&c)!=EOF){int i,j,n,l,r;for(i=1;i<=a;i++)scanf("%d",&st1[i]);for(i=1;i<=b;i++)scanf("%d",&st2[i]);    k=0;for(i=1;i<=a;i++){for(j=1;j<=b;j++){    k++;sum[k]=st1[i]+st2[j];}}sort(sum+1,sum+k+1);for(i=0;i<c;i++)scanf("%d",&st3[i]);scanf("%d",&n);t++;printf("Case %d:\n",t);while(n--){    scanf("%d",&x);            bool flag=true;            for(i=0;i<c;i++)            {            if(judge(x-st3[i]))            {            printf("YES\n");            flag=false;                break;}            }            if(flag)            printf("NO\n");}}}

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