hdoj 2141 Can you find it?【二分查找+暴力】

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 17036    Accepted Submission(s): 4337

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：给你三组数，每组数的个数分别为L N M再给你S个数x判断能否从这三组数中分别找到一个数使Ai+Bj+Ck = X.

题解：暴力肯定超时，所以要用二分，先将三组数中任意两组存入数组str[]中，然后再对每个x进行遍历

1、测试数据太弱，而且这样的一组测试数据本来就是个坑，好多不正确的情况都能得出正确答案

2、输出格式好坑，本来我以为是连续输出NO YES NO一直错没想到竟然是输一个数出一个结果

3、一定要注意给str[]数组排序（二分查找的条件是有序数列）

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#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define MAX 1100 bool cmp(int a,int b) {     return a<b; } int a[MAX]; int b[MAX]; int c[MAX]; int d,str[300000]; int main() {     int n,m,j,i;     int k=1,l;     while(scanf("%d%d%d",&l,&n,&m)!=EOF)     {         for(i=0;i<l;i++)             scanf("%d",&a[i]);             for(i=0;i<n;i++)                 scanf("%d",&b[i]);                 for(i=0;i<m;i++)                     scanf("%d",&c[i]);         int len=0;         for(i=0;i<n;i++)         {             for(j=0;j<m;j++)             {                 str[len++]=b[i]+c[j];             }         }         sort(str,str+len,cmp);         int ok;         printf("Case %d:\n",k++);         int s;         scanf("%d",&s);         for(i=0;i<s;i++)         {             scanf("%d",&d);             ok=0;             for(j=0;j<l;j++)             {                 int left = 0,right = len,mid = 0;                 int goal = d - a[j];                    while(right >= left)                 {                     mid = (right + left) >> 1;                     if(str[mid] < goal)                     left = mid + 1;                     else if(str[mid] > goal)                     right = mid - 1;                     else                     {                         ok=1;                         break;                     }                 }                 if(ok)                     break;             }             if(ok)                 printf("YES\n");             else                 printf("NO\n");         }     }     return 0; }