codeforces 712C C. Memory and De-Evolution(贪心)

题目链接：

C. Memory and De-Evolution

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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

input

8 5

output

input

22 4

output

6

题意：

把变成为x的等边三角形变成边长为y的三角形,其中过程要求三角形恒成立,最少需要操作多少次;

思路：

反过来贪心;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>using namespace std;#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));typedef  long long LL;template<class T> void read(T&num) {char CH; bool F=false;for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {if(!p) { puts("0"); return; }while(p) stk[++ tp] = p%10, p/=10;while(tp) putchar(stk[tp--] + '0');putchar('\n');
}const int mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=(1<<20)+10;
const int maxn=1e5+110;
const double eps=1e-12;int x,y;
queue<int>qu;
int main()
{read(x);read(y);int ans=0;qu.push(y);qu.push(y);qu.push(y);while(1){int f=qu.front();qu.pop();int g=qu.front();qu.pop();int u=qu.front();qu.pop();if(f>=x&&g>=x&&u>=x)break;ans++;qu.push(g);qu.push(u);qu.push(g+u-1);}cout<<ans<<endl;return 0;
}

转载于:https://www.cnblogs.com/zhangchengc919/p/5876913.html

codeforces 712C C. Memory and De-Evolution(贪心)相关推荐

codeforces#324(div2) E. Anton and Ira 贪心
codeforces#324(div2) E. Anton and Ira 贪心 E. Anton and Ira time limit per test 1 second memory limit ...
Codeforces 712C. Memory and De-Evolution
题目链接:http://codeforces.com/problemset/problem/712/C 题意: 给你两个值 a 和 b (a > b), 代表这里有两个等边三角形, 边长分别为 ...
Codeforces Round #192 (Div. 1) A. Purification 贪心
A. Purification Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/329/probl ...
Codeforces Beta Round #11 A. Increasing Sequence 贪心
A. Increasing Sequence 题目连接: http://www.codeforces.com/contest/11/problem/A Description A sequence a ...
Codeforces Round #382 (Div. 2)B. Urbanization 贪心
B. Urbanization 题目链接 http://codeforces.com/contest/735/problem/B 题面 Local authorities have heard a l ...
【Codeforces】925A Stairs and Elevators【贪心】
[Codeforces]925A Stairs and Elevators [题目大意] 在一个n*m的矩阵里,有clcl个楼梯和cece个电梯,电梯和楼梯可以到任意一层,给出clcl个楼梯的位置和c ...
Codeforces 814D An overnight dance in discotheque 贪心
题意题解 Codeforces 814D An overnight dance in discotheque 物语场!!!!之后的可怜和命运石之门--哈哈哈哈哈哈哈-- 题意平面直角坐标系中有n个 ...
Codeforces Round #579 (Div. 3) E. Boxers (贪心）
题目链接:https://codeforces.com/contest/1203/problem/E 思路:一开始想的是记录数字出现次数,看每个数字的三种变化能否产生新数,但是这样容易多算或者少算答案 ...
codeforces 584 E. Anton and Ira（贪心，数组p经变换到数组s）
题目:http://codeforces.com/contest/584/problem/E 题意:有两个数组p,s,长度为n,且是1~n的排列.要使数组p变为s,每交换 i 和 j 两个位子上的数, ...
【CodeForces 1100F】异或线性基 | 贪心 | 离线区间最大异或和 | E
补题ing- 真的不应该错过这场比赛的 qwq 1100F. Ivan and Burgers time limit per test: 3 seconds memory limit per test ...

codeforces 712C C. Memory and De-Evolution(贪心)

codeforces 712C C. Memory and De-Evolution(贪心)相关推荐

最新文章

热门文章