Problem Description

As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below:

A consists of integers between X and Y (inclusive).
For each 1≤i≤|A|?1, Ai+1 is a multiple of Ai and strictly greater than Ai.
Find the maximum possible length of the sequence.

Constraints

• 1≤X≤Y≤1018
• All input values are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the maximum possible length of the sequence.

Example

Sample Input 1

3 20

Sample Output 1

3
The sequence 3,6,18 satisfies the conditions.

Sample Input 2

25 100

Sample Output 2

3

Sample Input 3

314159265 358979323846264338

Sample Output 3

31

題意：給出 x、y 兩個數，在 x~y 范圍內構造一個序列，要求第 a[i+1] 個數是第 i 個數的倍數，求序列最大的可能的長度

思路：

通過題意可知，按：x、2x、4x、8x、16x、...、2^(i-1) x 的規則構造的序列一定是最長的

因此，只要枚舉 i 求出第一個大于?2^(i-1) x 的數時，i-1 就是答案

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 8000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;int main(){LL x,y;scanf("%lld%lld",&x,&y);for(LL i=0;i<100;i++){LL temp=pow(2,i-1)*x;if(temp>y){printf("%lld\n",i-1);break;}}return 0; }

?

總結

以上是生活随笔為你收集整理的Multiple Gift（AtCoder-3731）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。