这是 meelo 原创的 IEEEXtreme极限编程大赛题解

Xtreme 10.0 - Inti Sets

题目来源 第10届IEEE极限编程大赛

https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/inti-sets

In order to motivate his Peruvian students, a teacher includes words in the Quechua language in his math class.

Today, he defined a curious set for a given positive integer N. He called this set, an Inti set, and defined it as the set of all positive integer numbers that have the number 1 as their single common positive divisor with number N.

The math class about Inti sets was amazing. After class, the students try to challenge to teacher. They each ask questions like this: "Could you tell me the sum of all numbers, between A and B (inclusive), that are in the Inti set of N?"

Since the teacher is tired and he's sure that you are the best in class, he wants to know if you can help him.

Input Format

The first line of input contains an integer Q, 1 ≤ Q ≤ 20, representing the number of students. Each of the next Qlines contain three space-separated integers NA and B, which represent a query.

Constraints

1 ≤ A ≤ B ≤ N ≤ 10^12

Output Format

The output is exactly Q lines, one per student query. For each query you need to find the sum of all numbers between A and B, that are in the Inti set of N, and print the sum modulo 1000000007.

Sample Input

2
12 5 10
5 1 4

Sample Output

12
10

Explanation

In the sample input, Q = 2, so you have to answer two questions:

In the first question N = 12, A = 5 and B = 10. So you have to find the sum of all numbers between 5 and 10, that are in the Inti set of 12.

Inti set ( 12 ) = { 1, 5, 7, 11, 13, ... }

2 and 4 are not in the Inti set (12) because 12 and these numbers are also divisible by 2.

3 and 9 are not in the Inti set (12) because 12 and these numbers are also divisible by 3.

The numbers in the Inti set, which are in the query's range, are 5 and 7, so answer is ( 5 + 7 ) MOD 1000000007 = 12

In the second question, the numbers in the Inti set of 5 between 1 and 4 are: 1, 2, 3, 4; so the answer is ( 1 + 2 + 3 + 4 ) MOD 1000000007 = 10

题目解析

显然直接求和会超时,可以用容斥原理解决。

用sumOver(5, 10, 1)表示区间[5,10]内为1倍数的数

由于12的质因数为2, 3

sum(区间[5, 10]内与12互质的数) = sumOver(5, 10, 1) - sumOver(5, 10, 2) - sumOver(5, 10, 3) + sumOver(5, 10, 6)

可以通过遍历区间[0,2^2)的每一个数来遍历所有因式的组合,

二进制数形式每一位代表是否存在该因数,1代表存在,0代表不存在,

因数的个数为偶数意味着和需要加上,为奇数意味着需要减去

00代表因数为1, 01代表因数为3, 10代表因数为2, 11代表因数为6

注意需要使用取余运算避免溢出。

复杂度分析

如果有c个质因数,那么需要求2^c个数的和

求每一个和需要常数时间O(1)

数N,至多只有一个大于sqrt(N)的质因数,因此质因数的个数不超过log(sqrt(N))+1

总复杂度为O(sqrt(N))

程序

C++

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std; #define MAXN 1000000007 // 区间[a,b]内,所有为x倍数数的和
long long sumOver(long long a, long long b, long long x) {
long long aa = (a + x - ) / x; // 上取整
long long bb = b / x; // 下取整 long long sum; // sum会超过long long的表示范围
if( (aa + bb) % == ) {
sum = (((aa + bb) / ) % MAXN) * ((bb - aa + ) % MAXN);
} else {
sum = ((aa + bb) % MAXN) * (((bb - aa + ) / ) % MAXN);
} return ((sum % MAXN) * (x % MAXN)) % MAXN;
} // 求不大于max的所有素数
// 使用筛选法
void getPrimes(vector<long long> &primes, long long max) {
vector<bool> nums(max, );
for(long long i=; i<max; i++) {
if(nums[i] == false) {
primes.push_back(i);
for(int n=*i; n<max; n+=i) {
nums[n] = true;
}
}
}
} // 对数x进行质因数分解
void getFactors(long long x, vector<long long> &factors, vector<long long> &primes) {
int i = ;
while(x > && i < primes.size()) {
if(x % primes[i] == ) {
factors.push_back(primes[i]);
while(x % primes[i] == ) x /= primes[i];
}
i++;
}
// 小于10^12的数最对有一个大于10^6的质因数
if(x > ) {
factors.push_back(x);
}
} int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
vector<long long> primes;
getPrimes(primes, ); int T;
cin >> T;
for(int t=; t<T; t++) {
long long x, a, b;
cin >> x >> a >> b;
long long result = ;
vector<long long> factors;
getFactors(x, factors, primes); int factorCount = factors.size();
long long binMax = (long long) << factorCount; // 遍历所有的质因数组合
for(long long bin=; bin<binMax; bin++) {
long long factor = ;
int factorC = ;
for(int i=; i<factorCount; i++) {
if( (bin >> i) & ) {
factor *= factors[i];
factorC ++;
}
} if(factorC % == ) {
result = (result + sumOver(a, b, factor) + MAXN) % MAXN;
}
else {
result = (result - sumOver(a, b, factor) + MAXN) % MAXN;
}
} cout << result << endl;
} return ;
}

博客中的文章均为 meelo 原创,请务必以链接形式注明 本文地址

IEEEXtreme 10.0 - Inti Sets的更多相关文章

  1. IEEEXtreme 10.0 - Painter's Dilemma

    这是 meelo 原创的 IEEEXtreme极限编程比赛题解 Xtreme 10.0 - Painter's Dilemma 题目来源 第10届IEEE极限编程大赛 https://www.hack ...

  2. IEEEXtreme 10.0 - Ellipse Art

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Ellipse Art 题目来源 第10届IEEE极限编程大赛 https://www.hackerrank ...

  3. IEEEXtreme 10.0 - Counting Molecules

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Counting Molecules 题目来源 第10届IEEE极限编程大赛 https://www.hac ...

  4. IEEEXtreme 10.0 - Checkers Challenge

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Checkers Challenge 题目来源 第10届IEEE极限编程大赛 https://www.hac ...

  5. IEEEXtreme 10.0 - Game of Stones

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Game of Stones 题目来源 第10届IEEE极限编程大赛 https://www.hackerr ...

  6. IEEEXtreme 10.0 - Playing 20 Questions with an Unreliable Friend

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Playing 20 Questions with an Unreliable Friend 题目来源 第1 ...

  7. IEEEXtreme 10.0 - Full Adder

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Full Adder 题目来源 第10届IEEE极限编程大赛 https://www.hackerrank. ...

  8. IEEEXtreme 10.0 - N-Palindromes

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - N-Palindromes 题目来源 第10届IEEE极限编程大赛 https://www.hackerra ...

  9. IEEEXtreme 10.0 - Mysterious Maze

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Mysterious Maze 题目来源 第10届IEEE极限编程大赛 https://www.hacker ...

随机推荐

  1. 【arc075F】Mirrored

    Portal --> arc075_f Solution ​  一开始抱着"我有信仰爆搜就可以过"的心态写了一个爆搜.. ​  但是因为..剪枝和枚举方式不够优秀愉快T掉了q ...

  2. 怎样才能高效地使用JQuery

    1. 使用最新版本的jQuery jQuery的版本更新很快,你应该总是使用最新的版本.因为新版本会改进性能,还有很多新功能.下面就来看看,不同版本的jQuery性能差异有多大.这里是三条最常见的jQ ...

  3. laravel cookie加密解密原理

    通过控制台的 cookie 信息我们会发现,每次请求之后,关键的 cookie,如PHPSESSID.XSRF-TOKEN 都会发生变化,并且都是很长的一串字符串. 其实这是一个 json 数组,其中 ...

  4. 「Python」5个开源项目

    1-OpenAI universe Universe是一个能在世界上所有的游戏.网站和其他应用中,衡量和训练AI通用智能的软件平台. Universe,AI代理通过称为虚拟网络计算或VNC发送模拟的鼠 ...

  5. scrapy 采集网页出现丢失url的问题

    url_list = ["http://www.icoat.cc/news/list_18_3.html", "http://www.icoat.cc/news/list ...

  6. OpenCV---ROI(region of interest)和泛洪填充

    一:ROI 感兴趣区(Region of Interest,ROIs) 是图像的一部分,它通过在图像上选择或使用诸如设定阈值(thresholding) 或者从其他文件(如矢量> 转换获得等方法 ...

  7. [JSOI2010]Group

    1821: [JSOI2010]Group 部落划分 Group Time Limit: 10 Sec  Memory Limit: 64 MB[Submit][Status][Discuss] De ...

  8. 树链剖分处理+线段树解决问题 HDU 5029

    http://acm.split.hdu.edu.cn/showproblem.php?pid=5029 题意:n个点的树,m次操作.每次操作输入L,R,V,表示在[L,R]这个区间加上V这个数字.比 ...

  9. synchronized的实现原理

    常见三种使用方法: 1)普通同步方法,锁是当前实例: 2)静态同步方法,锁是当前类的Class实例,Class数据存在永久代中,是该类的一个全局锁: 3)对于同步代码块,锁是synchronized括 ...

  10. (三)配置Hadoop1.2.1+eclipse(Juno版)开发环境,并运行WordCount程序

    配置Hadoop1.2.1+eclipse(Juno版)开发环境,并运行WordCount程序 一.   需求部分 在ubuntu上用Eclipse IDE进行hadoop相关的开发,需要在Eclip ...