HDU 6069 Counting Divisors(求因子)
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Counting Divisors
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 2316 Accepted Submission(s): 833
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
31 5 11 10 21 100 3
Sample Output
10482302
Source
2017 Multi-University Training Contest - Team 4
题意:
d(n)=n的因子数。 给你l r k,求如上的表达式的值。
POINT:
如何求每个(l-r)^k的因子我就不说了。
之前遍历每个l-r求出各个的因子数会TLE。
那么遍历他们可能存在的质因数,即从2 3 5……开始遍历。
遍历后的质数就不用考虑了,在过程中保存下答案就行了。
还有一个重点就是,大质数的因子数是k+1没错,但是要考虑到这个大质数有可能是别的树除下来得到的,所以要更新sum数组。这个在代码里注释。//大质数即大于1e6即没有打到表的。
思维还不够灵活,总是没有想到从反面考虑。
#include <stdio.h>#include <iostream>#include <string.h>#include <math.h>#include <algorithm>#include <map>using namespace std;#define LL long longconst LL MAXN=1e6+9;const LL p = 998244353;LL prime[MAXN+1];LL sum[MAXN],num[MAXN];void getPrime(){ memset(prime,0,sizeof(prime)); for(int i=2;i<=MAXN;i++) { if(!prime[i])prime[++prime[0]]=i; for(int j=1;j<=prime[0]&&prime[j]<=MAXN/i;j++) { prime[prime[j]*i]=1; if(i%prime[j]==0) break; } }}int main(){ getPrime(); int cnt=(int)prime[0]; int T; scanf("%d",&T); while(T--) { LL l,r,k; scanf("%lld %lld %lld",&l,&r,&k); for(LL i=l;i<=r;i++) { sum[i-l]=1; num[i-l]=i; } for(int i=1;i<=cnt;i++) { //求出l-r内第一个可以整除prime[i]的。 LL flag=l%prime[i]; LL fir; if(!flag) fir=l; else fir=(l-flag)+prime[i]; for(LL j=fir;j<=r;j+=prime[i]) { int ci=0; while(num[j-l]%prime[i]==0) ci++,num[j-l]/=prime[i]; (sum[j-l]*=ci*k+1)%=p; } } LL ans=0; for(LL i=l;i<=r;i++) { if(num[i-l]!=1) sum[i-l]=(sum[i-l]*(k+1))%p;//大质数不绝对是k+1,sum[i-l]可能大于1.这个wa了蛮久的. (ans+=sum[i-l])%=p; } printf("%lld\n",ans); } return 0;}
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