HDU2141【hash】

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

 

分析：

用10^4打表  用10^2枚举

我的lower_bound一直挂  只能手写二分  不过还好

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 505;
int a[maxn], b[maxn], c[maxn], d[maxn * maxn];

int cnt;
bool check(int num) {
    int low = 0; int high = cnt - 1;
    while(low <= high) {
        int mid = ( low + high) >> 1;
        if(d[mid] >= num) {
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    if(d[high + 1] == num) return true;
    return false;
}

int main() {
    int l, n, m;
    int t = 1;
    while(EOF != scanf("%d %d %d",&l, &n, &m) ) {
        for(int i = 0; i < l; i++) scanf("%d",&a[i]);
        for(int i = 0; i < n; i++) scanf("%d",&b[i]);
        for(int i = 0; i < m; i++) scanf("%d",&c[i]);
        cnt = 0;
        for(int i = 0; i < l; i++) {
            for(int j = 0; j < n; j++) {
                d[cnt++] = a[i] + b[j];
            }
        }
        sort(d, d + cnt);
        int s;
        int num;
        printf("Case %d:\n", t++);
        scanf("%d",&s);
        while(s--) {
            scanf("%d",&num);
            bool flag = false;
            for(int i = 0; i < m; i++) {
                if(check(num - c[i]) ) {
                    flag = true;
                    break;
                }
            }
            if(flag) puts("YES"); 
            else puts("NO");
        }
    }
}