Co-prime 第一发容斥,感觉挺有意思的 →_→ [题目链接]Co-prime [题目类型]容斥 &题意: 求(a,b)区间内,与n互质的数的个数. \(a,b\leq 10^{15}\) &题解: 分析:我们可以先转化下:用(1,b)区间与n互质的数的个数减去(1,a-1)区间与n互质的数的个数,那么现在就转化成求(1,m)区间于n互质的数的个数,如果要求的是(1,n)区间与n互质的数的个数的话,我们直接求出n的欧拉函数值即可,可是这里是行不通的!我们不妨换一种思路:就是求出(1,m…

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,…

http://acm.hdu.edu.cn/showproblem.php?pid=5072 题意:给出N个数,求有多少个三元组,满足三个数全部两两互质或全部两两不互质. 题解: http://dtyfc.com/acm/980 我看的这个学会的…… 可以先求不满足要求的三元组数量,也就是abc,a和b互质,b和c不互质. 这样就要找这n个数中,和某个数不互质的数的个数. 可以质因数分解+容斥原理,求出和某个数不互质的数的个数(也就是和这个数有相同因数的数的个数). 还要先预处理以某个数x为因子…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1509    Accepted Submission(s): 592 Problem Description There are n people standing in a line. Each of them has a unique id number. Now the Ragn…

G - Coprime Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5072 Description There are n people standing in a line. Each of them has a unique id number. Now the Ragnarok is coming. We should ch…

Coprime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 130    Accepted Submission(s): 59 Problem Description There are n people standing in a line. Each of them has a unique id number. Now t…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4135 Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1176    Accepted Submission(s): 427 Problem Description Given a number N, you are a…

Coprime Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 849    Accepted Submission(s): 232 Problem Description Please write a program to calculate the k-th positive integer that is coprime with…

题目链接:Coprime pid=5072"> 题面: Coprime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1181    Accepted Submission(s): 471 Problem Description There are n people standing in a line. Each of t…

题目求[A,B]区间内与N互质数的个数. 可以通过求出区间内与N互质数的个数的前缀和,即[1,X],来得出[A,B]. 那么现在问题是求出[1,X]区间内与N互质数的个数,考虑这个问题的逆问题:[1,X]区间内与N不互质数的个数. 于是就可以先处理出N的所有质因数{p0,p1,p2,...,pn}. 而[1,X]能被pi整除的数有$\lfloor \frac X{p_i} \rfloor$个,再利用容斥原理除掉质因数公倍数重复计数的部分就能求出不互质个数. 最后X减去不互质个数就是互质个数了.…