Niuke algorithm Zhou Zhoulian 2-perfect square (play table, lower_bound, upper_bound)

 

Link: https://ac.nowcoder.com/acm/contest/5203/C
Source: Niuke.com

Time limit: C / C ++ 1 second, 2 seconds languages other
space restrictions: C / C ++ 131072K, other languages 262144K
64bit the IO the Format: LLD%

Title description

Query the number of perfect squares in the range of [l, r] multiple times

Define the integer x as a perfect square if and only if the integer y can be found such that y * y = x

Enter description:

A number n in the first line indicates the number
of queries after the number of queries.

Output description:

For each query, output a number to indicate the answer

Example 1

Input

5
1 3
1 4
2 4
4 4
1 1000000000

Output

1
2
1
1
31622

Remarks:

n <= 100000
0<= l <= r <= 1000000000

 

 

First pre- process all the total square numbers in [0,1000000000] , a total of 31,623 (0 ² ~ 31622 ² ).

Binary search finds  the position index of the first perfect square greater than or equal to  l . Binary search finds the position index of the first perfect square greater than r . Subtracting it is the answer. 

You can use lower_bound () and upper_bound () respectively

 

 1 #include <bits/stdc++.h>
 2 typedef long long LL;
 3 #define pb push_back
 4 const int INF = 0x3f3f3f3f;
 5 const double eps = 1e-8;
 6 const int mod = 1e9+7;
 7 const int maxn = 1e5+10;
 8 using namespace std;
 9 
10 vector<int> vt;
11 
12 int main()
13 {
14     #ifdef DEBUG
15     freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout);
16     #endif
17     
18     for(int i=0;i*i<=1000000000;i++)
19     vt.pb(i*i);
20 
21     int T;
22     scanf("%d",&T);
23     while(T--)
24     {
25         int l,r;
26         scanf("%d %d",&l,&r);
27         int pos1=lower_bound(vt.begin(),vt.end(),l)-vt.begin();
28         int pos2=upper_bound(vt.begin(),vt.end(),r)-vt.begin();
29         printf("%d\n",pos2-pos1);
30     }
31     
32     return 0;
33 }

 

 

 

 

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