hdu2141 二分查找

                **Can you find it?**

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 30503 Accepted Submission(s): 7594

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

题目分析，首先想到的是 暴力写，三个for循环，时间复杂度为n^3，对于题目的数据来说 显然是TLE的，之后就想到二分；
具体是这样做的，枚举 Ai+Bj,之后二分查找X-Ck

代码如下：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>#include<map>using namespace std;const int MAX = 510;int a[MAX],b[MAX],c[MAX],d[MAX*MAX];int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int l,n,m;int find(int l,int r,int key){    int mid;    while(l<=r){        mid = (l+r)/2;        if(d[mid] == key)            return 1;        else if(d[mid]<key)            l = mid+1;        else            r = mid-1;    }    return 0;}int main(void){    int cnt = 0;    while(scanf("%d %d %d",&l,&n,&m) != EOF){        int dlen = 0;        for(int i = 0 ; i < l;i++)            a[i] = read();        for(int i = 0 ; i < n;i++)            b[i] = read();        for(int i = 0 ; i < m;i++)            c[i] = read();        for(int i = 0 ; i < l ;i++){            for(int j = 0 ; j < n;j++)                d[dlen++] = a[i] + b[j];        }        sort(d,d+dlen);        int x;        x = read();        printf("Case %d:\n",++cnt);        for(int k = 0 ; k < x ;k++){            int y;            int f=1;            y = read();            for(int i = 0 ;i < m;i++){                f = 1;                if(find(0,dlen-1,y-c[i])){                    printf("YES\n");                    f = 0 ;                    break;                }            }            if(f) `              printf("NO\n");        }    }    return 0;}