題干：

Consider some square matrix?A?with side?n?consisting of zeros and ones. There are?nrows numbered from?1?to?n?from top to bottom and?n?columns numbered from?1?to?n?from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the?i-row and the?j-th column as?Ai,?j.

Let's call matrix?A?clear?if no two cells containing ones have a common side.

Let's call matrix?A?symmetrical?if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair?(i,?j)?(1?≤?i,?j?≤?n)?both of the following conditions must be met:?Ai,?j?=?An?-?i?+?1,?j?and?Ai,?j?=?Ai,?n?-?j?+?1.

Let's define the?sharpness?of matrix?A?as the number of ones in it.

Given integer?x, your task is to find the smallest positive integer?n?such that there exists a clear symmetrical matrix?A?with side?n?and sharpness?x.

Input

The only line contains a single integer?x?(1?≤?x?≤?100) — the required sharpness of the matrix.

Output

Print a single number — the sought value of?n.

Examples

Input

4

Output

3

Input

9

Output

5

Note

The figure below shows the matrices that correspond to the samples:

?

解題報告：

? ? 呵呵，打表找規(guī)律，首先可以發(fā)現(xiàn)n為偶數(shù)一定不行（比賽時1minAC的神仙不知道是怎么看出來的）

? ? 然后，除了3這個數(shù)，其他的都可以找個規(guī)律填數(shù)就可以了。

AC代碼：

#include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll x; int main() {int biao[25] = {0};cin>>x;for(int i = 1; i<=20; i++) {int now = (i+1)/2;if(i%2 == 0) biao[i] = biao[i-1];else biao[i] = now*now + (now-1)*(now-1);} // for(int i = 1; i<=17; i++) printf("%d\n",biao[i]);if(x== 3) printf("5\n");else printf("%d\n",lower_bound(biao+1,biao+20+1,x) - biao);return 0 ;}

?還未看懂的證明：https://blog.csdn.net/qq_24451605/article/details/48677823

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