hdu4135題解 容斥

Problem Description

Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

 

Input

The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10¹⁵) and (1 <=N <= 10⁹).

 

Output

For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.

 

Sample Input

2 1 10 2 3 15 5

 

Sample Output

Case #1: 5 Case #2: 10

Hint

In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

首先感謝ACM不懈的追求大神的博客，本題解的代碼是在該大神的題解的基礎上完成的：http://www.cnblogs.com/jiangjing/archive/2013/06/03/3115470.html

題目大意：給定整數N，求出多少a到b區間（包含a，b,以下關於區間的描述都包括邊界），有多少個數字和N互質。

思路：因爲數量巨大，如果用輾轉相除法會造成超時。我們應該回歸互質的定義：兩個數字的最大公約數爲1稱之爲互質。故兩數互質和兩數不互質（印象裏面好像叫做互餘吧）互爲獨立事件（這說話的腔調真是受不了。。。）。總而言之，如果我們能求出該區間有多少個數字和N不互質就好了。

那麼怎麼求不互質的個數呢？（設問既有過渡的功能也有引起讀者興趣的作用，哈哈，好了不扯了），好的，我們先把問題改成求1到b區間和N不互質的個數，我們要先找出N的所有質因數，並求出這些質因數相乘可以組合成哪些數字，並求出b除以這些數字的和（不保留小數部分），並且因爲有重疊的部分，我們這時候就需要用到互斥定理，例如:b=10,N=48，那麼質因數爲2，3。這些質因數相乘的組合有：1，2，3，6（可以選擇只要其中一個數甚至不選），這時候用互斥定理，答案就是：int（10/-1）+int（10/2）+int（10/3）+int（10/-6）,同理，我們可以求出1到a-1區間和N不互質的個數。

最難的骨頭啃下來了，我們可以求出答案，假設1到b區間和N不互質的個數爲f(b),假設1到a-1區間和N不互質的個數爲f(a-1)，那麼題目的答案就是b-f(b)-(a-1-f(a-1)),好的，此處應該有掌聲.

//hud 4135
#include<cstdio>
#include<iostream>
#include<vector>
#define maxn 100005
using namespace std;
vector <__int64> factor;
__int64 p[maxn];
int len_p;
void getfactor(__int64 x)
{
    factor.clear();
    for(__int64 i = 2; i * i <= x; i ++)
    {
        if(x % i == 0)
        {
            factor.push_back(i);
            while(x % i == 0)
            {
                x /= i;
            }
        }
    }
    if(x > 1)
    {
        factor.push_back(x);
    }
}
void getp()
{
    p[0] = -1;
    len_p = 1;
    int len_factor = factor.size();
    for(__int64 i = 0; i < len_factor; i ++)
    {
        int k = len_p;
        for(__int64 j = 0; j < k; j ++)
        {
            p[len_p ++] = factor[i] * p[j] * (-1);
        }
    }
}
__int64 f(__int64 x)
{
    __int64 ans = 0;
    for(int i = 1; i < len_p; i ++)
    {
        ans += (x / p[i]);
    }
    return ans;
}
int main()
{
    int d,cas = 1;
    cin >> d;
    while(d --)
    {
        __int64 a, b, n;
        scanf("%I64d%I64d%I64d", &a, &b, &n);
        getfactor(n);
        printf("Case #%d: ", cas ++);
        getp();
        cout << b - f(b) - (a - 1 - f(a - 1)) << endl;
    }
    return 0;
}
/*
2
1 10 2
3 15 5
ans:5 10
*/