C. Memory and De-Evolution
time limit per test
memory limit per test
input
output
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
Input
The first and only line contains two integers x and y (3 ≤ y < x ≤ 100 000) — the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
Examples
input
6 3
output
4
input
8 5
output
3
input
22 4
output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do
.In the second sample test, Memory can do
.In the third sample test, Memory can do:
.
反过来思考怎么把边长为y的等边三角形变为变为边长为x的等边三角形,每次执行这样的操作，把最小的边变为其他两条边长的和减一
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define maxn 100005
using namespace std;
typedef long long ll;
int main(){
int x, y, ans = 0;
scanf("%d%d", &x, &y);
int a = y, b = y, c = y;
while(a != x || b != x || c != x){
if(a < x){
a = b + c - 1;
if(a > x)a = x;
ans++;
}
if(b < x){
b = a + c - 1;
if(b > x)b = x;
ans++;
}
if(c < x){
c = a + b - 1;
if(c > x)c = x;
ans++;
}
}
printf("%d\n", ans);
return 0;
}