hdu 6069 统计区间约数的个数 2017 Multi-University Training Contest - Team 4
2017-08-04 13:21
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Counting Divisors
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1720 Accepted Submission(s): 629
[align=left]Problem Description[/align]
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, given l,r
and k,
your task is to calculate the following thing :
(∑i=lrd(ik))mod998244353
[align=left]Input[/align]
The first line of the input contains an integer
T(1≤T≤15),
denoting the number of test cases.
In each test case, there are 3
integers l,r,k(1≤l≤r≤1012,r−l≤106,1≤k≤107).
[align=left]Output[/align]
For each test case, print a single line containing an integer, denoting the answer.
[align=left]Sample Input[/align]
3
1 5 1
1 10 2
1 100 3
[align=left]Sample Output[/align]
10
48
2302
首先我们要知道一个定理
一个数的约数的个数等于,唯一分解成质数的形式后,每一个指数加一后累乘的结果
本题就比较简单了,比赛的时候傻逼了
枚举10的六次方以内的质数,不超过八万好像,忘了,可以运行程序看看就知道了,反正不多
然后再区间内部,枚举质数的倍数,这样可以减少很多运算量,然后计算这个质数在这个倍数里面可以分解得到多少个,即上面说的指数
需要注意的是,分解后可能会有大于10的六次方的指数,最后把这些数另外处理一下即可
#include<math.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define LL long long
#define mod 998244353
#define MAXN 1000006
LL sum[MAXN],num[MAXN];
LL primer[MAXN];
LL flag[MAXN];
int cnt=0;
void get_primer()
{
flag[0]=flag[1]=1;
LL sqrtmaxn=(LL)sqrt((double)MAXN);
for(int i=2;i<MAXN;i++){
if(primer[i]==0){
flag[i]=flag[i-1];
primer[cnt++]=i;
if(i<sqrtmaxn)
for(int j=i*i;j<MAXN;j+=i)
primer[j]=1;
}
else
flag[i]=(flag[i-1]*i)%mod;
}
}
int main()
{
int T;
get_primer();
LL a,b,c,ans;
//freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%lld%lld%lld",&a,&b,&c);
for(int i=0;i<=b-a+1;i++)
sum[i]=1,num[i]=a+i;
LL pos,t,temp;
for(int i=0;i<cnt;i++){
if(primer[i]>(b/2+1))
break;
t=a/primer[i];
if(a%primer[i]) t++;
pos=t*primer[i];
for(LL j=pos;j<=b;j+=primer[i]){
t=0;
while(num[j-a]%primer[i]==0)
num[j-a]/=primer[i],t++;
sum[j-a]=sum[j-a]*(c*t+1)%mod;
}
}
ans=0;
for(LL j=a;j<=b;j++){
if(num[j-a]>1)
sum[j-a]=sum[j-a]*(c+1)%mod;
ans=(ans+sum[j-a])%mod;
}
printf("%lld\n",ans);
}
return 0;
}
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