HDU 6069 Counting Divisors【区间素筛】【经典题】【好题】
Counting Divisors
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1996 Accepted Submission(s): 728
Problem Description
In mathematics, the function d(n) denotes
the number of divisors of positive integer n .
For example,d(12)=6 because 1,2,3,4,6,12 are
all 12 's
divisors.
In this problem, givenl,r and k ,
your task is to calculate the following thing :
(∑i=lrd(ik))mod998244353
For example,
In this problem, given
Input
The first line of the input contains an integer T(1≤T≤15) ,
denoting the number of test cases.
In each test case, there are3 integers l,r,k(1≤l≤r≤1012,r−l≤106,1≤k≤107) .
In each test case, there are
Output
For each test case, print a single line containing an integer, denoting the answer.
Sample Input
3 1 5 1 1 10 2 1 100 3
Sample Output
10 48 2302
Source
题意:求l到r中所有数的k次方的因数个数之和。
题解:
设n=p1c1p2c2...pmcm,则d(nk)=(kc1+1)(kc2+1)...(kcm+1),因为区间长度不超过1e6枚举不超过
注意:这里分解的时候不是把每一个数逐一进行分解,而是类似于区间两次筛,每一次枚举素数的倍数,然后整段的去筛计算素数的指数,相当于每次找len/2,len/3,len/4...len/n次,以节约时间,与poj2689的思路有异曲同工之妙。
写的时候傻逼了,把
for (int j = 2 * i; j < maxn; j += i) not_prime[j] = 1;
的not_prime[j] = 1写成了not_prime[i] = 1,T了一万年 (T_T)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <cstring> #define INF 0x3f3f3f3f #define ms(x,y) memset(x,y,sizeof(x)) using namespace std; typedef long long ll; const double pi = acos(-1.0); const double eps = 1e-8; const int mod = 998244353; const int maxn = 1e6 + 10; ll prime[maxn]; //记录素数 ll l, r, k; ll d[maxn]; //记录这个数的质因数数量 ll ans[maxn]; //记录这个数是否可以被完全分解质因数 bool not_prime[maxn]; int cnt = 0; //素数的个数 void getPrime() //埃氏筛法得到1e6内的素数 { not_prime[0] = not_prime[1] = 1; for (int i = 2; i < maxn; i++) { if (!not_prime[i]) { prime[cnt++] = i; for (int j = 2 * i; j < maxn; j += i) not_prime[j] = 1; } } } void solve() { for (int i = 0; i < cnt; i++) //枚举素数 { ll tmp = (l + prime[i] - 1) / prime[i] * prime[i]; //利用自动取整【(l + prime[i] - 1) / prime[i]】得到l到r的最小的prime[i]的倍数,乘prime[i]得到这个数 while (tmp <= r) { ll cnt = 0; //记录指数是多少 while (ans[tmp - l] % prime[i] == 0) { cnt++; ans[tmp - l] /= prime[i]; } d[tmp - l] = (d[tmp - l] * (k*cnt + 1)) % mod; tmp += prime[i]; //枚举下一倍数 } } ll res = 0; for (ll i = l; i <= r; i++) { if (ans[i - l] == 1) res = (res + d[i - l]) % mod; //如果ans结果为1则证明可以完全分解,直接加上结果即可 else res = (res + d[i - l] * (k + 1)) % mod; //否则表示还有素数且素数指数为1,再乘以1*k+1即可 } printf("%lld\n", res); } int main() { getPrime(); int t; scanf("%d", &t); while (t--) { scanf("%lld%lld%lld", &l, &r, &k); for (ll i = l; i <= r; i++) //初始化d和ans { d[i - l] = 1; ans[i - l] = i; } solve(); } return 0; }
Fighting~