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HDU2141【hash】

时间:2014-10-30 20:42:05      阅读:139      评论:0      收藏:0      [点我收藏+]

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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

 

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:

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn = 505;
 8 int a[maxn], b[maxn], c[maxn], d[maxn * maxn];
 9 
10 int cnt;
11 bool check(int num) {
12     int low = 0; int high = cnt - 1;
13     while(low <= high) {
14         int mid = ( low + high) >> 1;
15         if(d[mid] >= num) {
16             high = mid - 1;
17         } else {
18             low = mid + 1;
19         }
20     }
21     if(d[high + 1] == num) return true;
22     return false;
23 }
24 
25 int main() {
26     int l, n, m;
27     int t = 1;
28     while(EOF != scanf("%d %d %d",&l, &n, &m) ) {
29         for(int i = 0; i < l; i++) scanf("%d",&a[i]);
30         for(int i = 0; i < n; i++) scanf("%d",&b[i]);
31         for(int i = 0; i < m; i++) scanf("%d",&c[i]);
32         cnt = 0;
33         for(int i = 0; i < l; i++) {
34             for(int j = 0; j < n; j++) {
35                 d[cnt++] = a[i] + b[j];
36             }
37         }
38         sort(d, d + cnt);
39         int s;
40         int num;
41         printf("Case %d:\n", t++);
42         scanf("%d",&s);
43         while(s--) {
44             scanf("%d",&num);
45             bool flag = false;
46             for(int i = 0; i < m; i++) {
47                 if(check(num - c[i]) ) {
48                     flag = true;
49                     break;
50                 }
51             }
52             if(flag) puts("YES"); 
53             else puts("NO");
54         }
55     }
56 }
View Code

 

HDU2141【hash】

标签:des   style   blog   http   io   color   os   ar   for   

原文地址:http://www.cnblogs.com/zhanzhao/p/4063532.html

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