Codeforces Round #392 (Div. 2) F. Geometrical Progression
原题地址:http://codeforces.com/contest/758/problem/F
F. Geometrical Progression
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
For given n, l and r find the number of distinct geometrical progression, each of which contains n distinct integers not less than l and not greater than r. In other words, for each progression the following must hold: l ≤ ai ≤ r and ai ≠ aj , where a1, a2, ..., an is the geometrical progression, 1 ≤ i, j ≤ n and i ≠ j.
Geometrical progression is a sequence of numbers a1, a2, ..., an where each term after first is found by multiplying the previous one by a fixed non-zero number d called the common ratio. Note that in our task d may be non-integer. For example in progression 4, 6, 9, common ratio is .
Two progressions a1, a2, ..., an and b1, b2, ..., bn are considered different, if there is such i (1 ≤ i ≤ n) that ai ≠ bi.
Input
The first and the only line cotains three integers n, l and r (1 ≤ n ≤ 107, 1 ≤ l ≤ r ≤ 107).
Output
Print the integer K — is the answer to the problem.
Examples
Input
1 1 10
Output
10
Input
2 6 9
Output
12
Input
3 1 10
Output
8
Input
3 3 10
Output
2
Note
These are possible progressions for the first test of examples:
- 1;
- 2;
- 3;
- 4;
- 5;
- 6;
- 7;
- 8;
- 9;
- 10.
These are possible progressions for the second test of examples:
- 6, 7;
- 6, 8;
- 6, 9;
- 7, 6;
- 7, 8;
- 7, 9;
- 8, 6;
- 8, 7;
- 8, 9;
- 9, 6;
- 9, 7;
- 9, 8.
These are possible progressions for the third test of examples:
- 1, 2, 4;
- 1, 3, 9;
- 2, 4, 8;
- 4, 2, 1;
- 4, 6, 9;
- 8, 4, 2;
- 9, 3, 1;
- 9, 6, 4.
These are possible progressions for the fourth test of examples:
- 4, 6, 9;
- 9, 6, 4.
题意:给定 n, l and r ,求项数为n, 公比不为1,且数列每一项都属于[l,r]范围的不同的 等比数列 的个数。
题解:其实是先缩小范围然后直接枚举。
考虑数据范围1 ≤ n ≤ 107, 1 ≤ l ≤ r ≤ 107
设等比数列公比为d, d表示为 q/p,其中q或p为不同时等于1,且互质的正整数。
递增和递减数列的情况是成对出现的,即p和q互换。
所以不妨只考虑递增数列的情况,即公比d表示为q/p,其中pq互质,p为任意正整数,q>p,q为大于等于2的正整数。
则数列末项整除于qn-1 ,其中q>=2,2^24>10^7, 故n>=24时无解。
n=1时为结果为r-l+1, n=2时结果为(r-l+1)*(r-l),n>24时0.
n>=3&&n<24时,可以通过枚举出p和q的情况求解。
n>=3, 由于数列末项整除于qn-1 ,则qn-1 ≤ 107,即枚举 p,q的上界是(107)1/(n-1),当n=3时,这个值为3162,可以通过暴力枚举实现。
枚举p,q,
每找到一对(p,q)且gcd(p,q)==1
考虑数列末项 an= a1*qn-1/pn-1 ,
要满足 a1>=l, an<=r 的范围条件,若 l*qn-1/pn-1 >r 则不满足题意,continue;
若 l*qn-1/pn-1 <=r 则有满足[l,r]范围的等比数列
现在求[l,r]范围,公比为q/p,项数为n的等比数列的个数。
数列各项为 a1, a1*q/p ……a1*qn-1/qn-1qn-1pn-1 /pn-1 /pn-1 pn-1q/pq/pq/pn-1 ,等比数列的个数即为a1可能的值。
末项为moa1*qn-1/ pn-1 所以a1必整除于pn-1 ,即a1可能的值为 [l,r*pn-1/qn-1]范围内可被 pn-1整除的, 即 (r*pn-1/qn-1)/pn-1-l/pn-1个
#include <bits/stdc++.h>
#define LL long long
using namespace std; LL gcd(LL a, LL b){
if(b==) return a;
else return gcd(b,a%b);
} LL QuickPow(LL a, LL n){
LL ret=;
while(n){
if(n&) ret*=a;
a*=a;
n>>=;
}
return ret;
} LL l,r,n;
LL ans; int main()
{
cin>>n>>l>>r;
if(n>){
cout<<;return ;
}
if(n==){
cout<<r-l+;return ;
}
if(n==){
cout<<(r-l+)*(r-l);return ;
} //n>=3&&n<24的情况
LL upperlimit,pn,qn;
//p,q的枚举上界
upperlimit=pow(,double(log2(1e7+)/(n-))); //注意精度
for(LL p=;p<=upperlimit;p++)
for(LL q=p+;q<=upperlimit;q++)
if(gcd(p,q)==)
{
qn=QuickPow(q,n-);
pn=QuickPow(p,n-);
if(l*qn/pn>r) continue; //a1可能的值 :[l,r*pn/qn]范围内可被 pn整除的正整数,
ans+=(r*pn/qn)/pn-(l-)/pn;
}
//递增数列递减数列成对出现,只考虑了递增数列
cout<<ans*;
return ;
}
a1*qn-1/qn-1qn-1pn-1 /pn-1 /pn-1 pn-1q/pq/pq/pn-1 qn-1/qn-1qn-1pn-1 /pn-1 /pn-1 pn-1q/pq/pq/pn的
Codeforces Round #392 (Div. 2) F. Geometrical Progression的更多相关文章
- Codeforces Round #485 (Div. 2) F. AND Graph
Codeforces Round #485 (Div. 2) F. AND Graph 题目连接: http://codeforces.com/contest/987/problem/F Descri ...
- Codeforces Round #486 (Div. 3) F. Rain and Umbrellas
Codeforces Round #486 (Div. 3) F. Rain and Umbrellas 题目连接: http://codeforces.com/group/T0ITBvoeEx/co ...
- Codeforces Round #501 (Div. 3) F. Bracket Substring
题目链接 Codeforces Round #501 (Div. 3) F. Bracket Substring 题解 官方题解 http://codeforces.com/blog/entry/60 ...
- map Codeforces Round #Pi (Div. 2) C. Geometric Progression
题目传送门 /* 题意:问选出3个数成等比数列有多少种选法 map:c1记录是第二个数或第三个数的选法,c2表示所有数字出现的次数.别人的代码很短,思维巧妙 */ /***************** ...
- Codeforces Round #499 (Div. 1) F. Tree
Codeforces Round #499 (Div. 1) F. Tree 题目链接 \(\rm CodeForces\):https://codeforces.com/contest/1010/p ...
- Codeforces Round #376 (Div. 2)F. Video Cards(前缀和)
题目链接:http://codeforces.com/contest/731/problem/F 题意:有n个数,从里面选出来一个作为第一个,然后剩下的数要满足是这个数的倍数,如果不是,只能减小为他的 ...
- Codeforces Round #271 (Div. 2) F. Ant colony (RMQ or 线段树)
题目链接:http://codeforces.com/contest/474/problem/F 题意简而言之就是问你区间l到r之间有多少个数能整除区间内除了这个数的其他的数,然后区间长度减去数的个数 ...
- Codeforces Round #325 (Div. 2) F. Lizard Era: Beginning meet in the mid
F. Lizard Era: Beginning Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
- Codeforces Round #271 (Div. 2) F题 Ant colony(线段树)
题目地址:http://codeforces.com/contest/474/problem/F 由题意可知,最后能够留下来的一定是区间最小gcd. 那就转化成了该区间内与区间最小gcd数相等的个数. ...
随机推荐
- 【微信】微信小程序 微信开发工具 创建js文件报错 pages/module/module.js 出现脚本错误或者未正确调用 Page()
创建报错pages/module/module.js 出现脚本错误或者未正确调用 Page() 解决方法: 在js文件中添加 Page({ })
- js延时函数setTimeout
实现一个延时执行的效果,现记录如下: <html> <head> <script type="text/javascript" src="/ ...
- docker下载ubuntu并进行修改后生成新的镜像提交
一 docker pull ubuntu ,先下载下来一个镜像, 或者 从本地启动一个镜像 docker run -i -t ubuntu /bin/bash 二 进入一定更新操作 # shell ...
- 一个java调用python的问题
使用 ProcessBuilder List<String> commands = new ArrayList(); commands.add("python"); c ...
- How to initialize th rasp berry PI
WHAT YOU WILL NEED REQUIRED SD Card We recommend an 8GB class 4 SD card – ideally preinstalled with ...
- Angular 学习笔记——$rounte
<!DOCTYPE HTML> <html ng-app="myApp"> <head> <meta http-equiv="C ...
- 系统重装 如何在PC上安装Mac OS,苹果操作系统
[苹果系统 无影精品]<精睿 无影技术Mac OS X 中文优化正式会员版V10.14和V10.16>◆ 系统前沿:==================================== ...
- 开发ionic准备之安卓模拟器设置(2)
发现这个安卓模拟器设置屏幕还不能太大,太大显示不全,然后整个模拟器不能拖动,所以尽量不要设置太大的分辨率 ,如下即可 如果选安卓4.4然后勾选了其他下面的ok还不能点击的话,这下要去sdk manag ...
- uva507 - Jill Rides Again(最长连续和)
option=com_onlinejudge&Itemid=8&page=show_problem&problem=448">题目:uva507 - Jill ...
- ExtJs的Ext.grid.GridPanel不能选择复制表格中的内容解决方案
今天遇到grid复制的问题,在网上找到了一个解决办法,只需改下CSS和JS,给大家分享一下: 本文来自CSDN博客,转载请标明出处:http://blog.csdn.net/dy_paradise/a ...