Can you find it? HDU-2141 （二分查找模版题）

短命女 2021-11-05 06:14 214阅读 0赞

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

思路：枚举三个数组，时间复杂度O(N^3)，肯定会超时。所以可以把前两个数组的和先枚举出来，存在数组sum\[260000\]中，

由于要找是否存在sum\[i\] + c\[j\] = X, 可以把问题转换为在sum数组中查找是否有X-c\[j\]这个数。用二分查找sum数组即可。时间复杂度为： S \* c数组的大小 \* log(sum数组的大小)

注意：不能在c数组中二分查找是否有X-sum\[i\]这个数，因为其时间复杂度为： S \* sum数组的大小 \* log(c数组的大小)，由于题目中sum数组大小最大为250000，c数组大小最大为500，S为1000，所以会超时

1 #include <iostream>
     2 #include <queue>
     3 #include <cstring>
     4 #include <cstdio>
     5 #include <string>
     6 #include <algorithm>
     7 
     8 
     9 using namespace std;
    10 
    11 int a[501], b[501], c[501];
    12 int sum[260000];
    13 int L, N, M, S, X;
    14 int cnt = 1;
    15 
    16 int main()
    17 {
    18     while(scanf("%d %d %d", &L, &N, &M) != EOF)
    19     {
    20         for(int i = 0; i < L; ++i)
    21             scanf("%d", &a[i]);
    22         for(int i = 0; i < N; ++i)
    23             scanf("%d", &b[i]);
    24         for(int i = 0; i < M; ++i)
    25             scanf("%d", &c[i]);
    26             
    27         int k = 0;
    28         for(int i = 0; i < L; ++i)
    29             for(int j = 0; j < N; ++j)
    30                 sum[k++] = a[i] + b[j];
    31         
    32         sort(sum, sum+k);
    33         scanf("%d", &S);
    34         printf("Case %d:\n", cnt++);
    35         while(S--)
    36         {
    37             scanf("%d", &X);
    38             
    39             int flag = 0;
    40             for(int i = 0; i < M; ++i)
    41             {
    42                 int target = X - c[i];
    43                 int left = 0, right = k;
    44                 
    45                 while(left <= right)
    46                 {
    47                     int mid = (left + right) / 2;
    48                     if(sum[mid] == target)
    49                     {
    50                         flag = 1;
    51                         break;
    52                     }
    53                     else if(sum[mid] < target)
    54                         left = mid + 1;
    55                     else
    56                         right = mid - 1;
    57                 }
    58                 if(flag == 1)
    59                 {
    60                     printf("YES\n");
    61                     break;
    62                 }
    63             }
    64             if(flag == 0)
    65                 printf("NO\n");
    66             
    67         }
    68     }
    69     
    70 
    71     
    72     return 0;
    73 }

转载于:https://www.cnblogs.com/FengZeng666/p/11302734.html