HDU2141 Can you find it?(二分)

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 30394    Accepted Submission(s): 7578

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

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest 

http://acm.split.hdu.edu.cn/showproblem.php?pid=2141

题意 给你三个数量分别为 L,N,M的序列   有S个数     分别从三个数组里面拿出一个 看三个数之和能不能为 这些数

如果暴力的话肯定是超时的 把两个数组相加得到500 *500 的数组 二分

我是借助SET实现的   好像 一下子内存一下子大了好多倍 不知道为什么 G++交没过 C++就过了

大概是这个原因吧  http://blog.csdn.net/xky140610205/article/details/75212155

#include<stdio.h>#include<set>using namespace std;set<int>ss;set<int>::iterator it;int casee;int a[555],c[555];int main(){    casee=1;    int l,j,n,m,i,dt,jl,sum;    while(scanf("%d %d %d",&l,&n,&m)!=EOF)    {        ss.clear();        for(i=1; i<=l; i++)            scanf("%d",&a[i]);        for(i=1; i<=n; i++)        {            scanf("%d",&jl);            for(j=1; j<=l; j++)            {                dt=jl+a[j];                ss.insert(dt);            }        }        for(i=1; i<=m; i++)            scanf("%d",&c[i]);        int s,gg;        printf("Case %d:\n",casee++);        scanf("%d",&s);        while(s--)        {            scanf("%d",&sum);            int flag=0;            for(j=1; j<=m; j++)            {                gg=sum-c[j];                it=ss.find(gg);                if(it!=ss.end())                {                    printf("YES\n");                    flag=1;                    break;                }            }            if(!flag) printf("NO\n");        }    }    return 0;}