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 DescriptionGive 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. 

InputThere 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. 

OutputFor 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

 

Authorwangye 

SourceHDU 2007-11 Programming Contest 

Recommend威士忌

解題思路：數組合并+二分尋找（搜尋），（解法一）
          這個搜尋讓額淚牛滿面啊，真的想先到用數組尋找，最近搜尋搜得吐蒙頭。。。。。。
        
          開始，搜尋一個數組得出x-d[i]的差值後，再在合并的數組裡面搜尋差值即可。

#include<stdio.h>#include<algorithm>using namespace std;#define K 505int LN[K*K];int BinarySearch(int LN[],int h,int t)/*二分尋找*/{    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); /*對LN數組排序*/        scanf("%d",&S);        printf("Case %d:\n",count++);        for(i=0;i<S;i++)        {            scanf("%d",&q);/*q即為題目中的x*/            int p=0; /*p為標記，0為找不到，1為能找到*/            for(j=0;j<m;j++)            {                int a=q-M[j]; /*因為L[i]+N[j]+M[k]==q，所以q-M[k]=LN[h]*/                if(BinarySearch(LN,h,a)) /*在LN數組中尋找到a*/                {                    printf("YES\n");                    p=1;                    break;                }            }            if(!p) /*找不到a*/              printf("NO\n");        }    }    return 0;}

解題思路：數組合并+set尋找（搜尋），（方法二）STL
         該解法，首先感謝蠶豆兒（http://blog.csdn.net/wqc359782004）給我靈感，他說可以用map搜尋，我才想到用STL的，後面發現map超記憶體了，才想起記憶體需求更小的set（map的1/2就好了）。
         開始，搜尋一個數組得出x-d[i]的差值後，再在合并的數組的set裡面搜尋差值即可。
                       附：STL函數方法集合（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作為尋找輔助using namespace std;int main(){    int b[500];    int c[500];    int d[500];    set< int> st;  //set定義，不能用map<int ,int>,記憶體翻倍，會記憶體    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++)    //數組合并            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]))  //值檢索，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;}