Codeforces Round #370 (Div. 2) C 贪心

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C. Memory and De-Evolution

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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: 

题意：

给出一个等边三角形的边长为x，要求每次改变一条边的边长，使得三角形尽可能的快速达到另外一个等边三角形，且边长为y，缩短边长的时候要保证当前的三条边能构成三角形

思路：

倒着从y->x贪心，每次把最小的边换成最大的

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<list>#include<stack>#include<iomanip>#include<cmath>#include<bitset>#define mst(ss,b) memset((ss),(b),sizeof(ss))///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;#define INF (1ll<<60)-1#define Max 1e9using namespace std;int n,m;int main(){    scanf("%d%d",&n,&m);    int a[4];    a[1]=a[2]=a[3]=m;    int ans=0;    while(1){        if(a[1]==n && a[2]==n && a[3]==n) break;        sort(a+1,a+4);        if(a[1]+a[3]<=a[2]+a[3]) a[1]=min(n,a[2]+a[3]-1);        else a[1]=min(n,a[1]+a[3]-1);        ans++;        //cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<endl;    }    cout<<ans<<endl;    return 0;}