HDU4135容斥原理递归与非递归版

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Co-prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1207 Accepted Submission(s): 440

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

21 10 23 15 5

Sample Output

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

Source

The Third Lebanese Collegiate Programming Contest

非递归版

#define DeBUG#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <set>#include <sstream>#include <map>#include <bitset>using namespace std ;#define zero {0}#define INF 0x3f3f3f3f#define EPS 1e-6typedef long long LL;const double PI = acos(-1.0);//#pragma comment(linker, "/STACK:102400000,102400000")inline int sgn(double x){    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}#define N 100005std::vector<int> vt;long long n, a, b, res;long long que[10240];void fx()//提取因子{    vt.clear();    res = n;    for (int i = 2; i * i <= n; i++)    {        if (res % i == 0)        {            vt.push_back(i);            while (res % i == 0)            {                res /= i;            }        }    }    if (res > 1)vt.push_back(res);}//比如求1~30 内与30互质的//30-30/2-30/3+30/6-30/5+30/10+30/15-30/30//que数组里就是 1,-2,-3,6,-5,10,15,-30long long cal(long long n, long long t){    int num = 0;    que[num++] = 1;    for (int i = 0; i < vt.size(); i++)    {        int ep = vt[i];        int k = num;        for (int j = 0; j < k; j++)            que[num++] = ep * que[j] * (-1);    }    long long sum = 0;    for (int i = 0; i < num; i++)        sum += t / que[i];    return sum;}int main(){#ifdef DeBUGs    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);#endif    int T, tcase = 0;    cin >> T;    while (T--)    {        cin >> a >> b >> n;        fx();        long long ans = cal(n, b) - cal(n, a - 1);        printf("Case #%d: %I64d\n", ++tcase, ans);    }    return 0;}

递归版，转自http://blog.csdn.net/dellaserss/article/details/8239952

不贴了。。。

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