HDU2141 Can you find it?

                                                       Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 7927    Accepted Submission(s): 2062

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
   
   
   
   

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

 

Recommend

威士忌



解题思路：数组合并+二分查找（搜索），（解法一）
          这个搜索让额泪牛满面啊，真的想先到用数组查找，最近搜索搜得吐蒙头。。。。。。
        
          开始，搜索一个数组得出x-d[i]的差值后，再在合并的数组里面搜索差值即可。

#include<stdio.h> #include<algorithm> using namespace std; #define K 505 int LN[K*K]; int BinarySearch(int LN[],int h,int t)/*二分查找*/ {     int left,right,mid;     left=0;     right=h-1;     mid=(left+right)/2;     while(left<=right)     {         mid=(left+right)/2;         if(LN[mid]==t)           return 1;         else if(LN[mid]>t)           right=mid-1;         else if(LN[mid]<t)           left=mid+1;     }     return 0; } int main() {     int i,j,count=1,q;     __int32 L[K],N[K],M[K],S,n,m,l;     while(scanf("%d%d%d",&l,&n,&m)!=EOF)     {         int h=0;         for(i=0;i<l;i++)          scanf("%d",&L[i]);         for(i=0;i<n;i++)           scanf("%d",&N[i]);         for(i=0;i<m;i++)           scanf("%d",&M[i]);         for(i=0;i<l;i++)           for(j=0;j<n;j++)            LN[h++]=L[i]+N[j];/*合并L和N*/         sort(LN,LN+h); /*对LN数组排序*/         scanf("%d",&S);         printf("Case %d:\n",count++);         for(i=0;i<S;i++)         {             scanf("%d",&q);/*q即为题目中的x*/             int p=0; /*p为标记，0为找不到，1为能找到*/             for(j=0;j<m;j++)             {                 int a=q-M[j]; /*因为L[i]+N[j]+M[k]==q，所以q-M[k]=LN[h]*/                 if(BinarySearch(LN,h,a)) /*在LN数组中查找到a*/                 {                     printf("YES\n");                     p=1;                     break;                 }             }             if(!p) /*找不到a*/               printf("NO\n");         }     }     return 0; }

解题思路：数组合并+set查找（搜索），（方法二）STL
         该解法，首先感谢蚕豆儿（http://blog.csdn.net/wqc359782004）给我灵感，他说可以用map搜索，我才想到用STL的，后面发现map超内存了，才想起内存需求更小的set（map的1/2就好了）。
         开始，搜索一个数组得出x-d[i]的差值后，再在合并的数组的set里面搜索差值即可。
                       附：STL函数方法集合（http://blog.csdn.net/lsh670660992/article/details/9204285），map用法（http://blog.csdn.net/lsh670660992/article/details/9158539），set用法（http://blog.csdn.net/lsh670660992/article/details/9158573）

#include<cstdio> #include<cstring> #include<algorithm> #include<set>    //用到STL中得set作为查找辅助 using namespace std; int main() {     int b[500];     int c[500];     int d[500];     set< int> st;  //set定义，不能用map<int ,int>,内存翻倍，会内存     int x;     int l,n,m;     int i,j,p=1;     while(scanf("%d%d%d",&l,&n,&m)!=EOF)     {         memset(b,0,sizeof(b));         memset(c,0,sizeof(c));         memset(d,0,sizeof(d));         for(i=0;i<l;i++)             scanf("%d",&b[i]);         for(i=0;i<n;i++)             scanf("%d",&c[i]);         for(i=0;i<m;i++)             scanf("%d",&d[i]);         for(j=0;j<n;j++)    //数组合并             for(i=0;i<l;i++)                 st.insert(b[i]+c[j]);   //值插入         printf("Case %d:\n",p++);         scanf("%d",&n);         while(n--)         {             int s;             scanf("%d",&x);             for(i=0;i<m;i++)                 if(st.count(x-d[i]))  //值检索，存在返回1，不存在返回0                 {                     printf("YES\n");                     break;                 }             if(i==m)                 printf("NO\n");         }         st.clear();     }     return 0; }