2017 Multi-University Training Contest
来源:互联网 发布:淘宝积分在哪里查看 编辑:程序博客网 时间:2024/06/05 12:44
题目:
Problem Description
In mathematics, the functiond(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 integerT(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
题意:求区间[l,r]中的元素i^k的因子数之和。
思路:由于数据过大,所以暴力是行不通的!!!根据定理(貌似叫约数定理):n=p1^a1×p2^a2×p3^a3*…*pm^am,n的因子数的个数就是(a1+1)(a2+1)(a3+1)…(am+1)。所以可以推理出n^k的因子数为(k*a1+1)(k*a2+1)(k*a3+1)…(k*am+1)。由于r最大值为1e12,所以它的质因子最大不超过1e6。把[1,1e6]的素数利用素数筛法求出。枚举每个素数在区间内的倍数,可以跳着枚举,计算出每个数对应的因子个数。若区间中的数不断模除因子后的最終结果不是1,则说明这个数本身是素数,素数的因子个数就等于枚举的因子个数*(k+1)累乘起来。。。吐槽:不知道为什么一直TLE、RE,然后去看别人的题解没找到错误,替换两行相同的代码后就AC了,真的醉了,玄学啊附上TLE代码一份,望大佬指教有何不同
CODE:
TLE:
#include<bits/stdc++.h>using namespace std;typedef long long LL;#define mod 998244353#define maxn 1000001bool isprime[maxn];int cnt;LL prime[maxn],f[maxn],num[maxn],l,r,k;void Prime(){ cnt=0; memset(isprime,true,sizeof(isprime)); isprime[1]=false; for(int i=2;i<maxn;i++){ if(isprime[i]){ prime[cnt++]=i; for(int j=2;j*i<maxn;j++) isprime[i*j]=false; } }}int main(){ Prime(); int t; scanf("%d",&t); while(t--){ scanf("%lld%lld%lld",&l,&r,&k); for(int i=0;i<=r-l;i++) f[i]=i+l,num[i]=1; for(int i=0;i<cnt;i++){ LL st=(l/prime[i]+(l%prime[i]?1:0))*prime[i]; for(LL j=st;j<=r;j+=prime[i]){ LL len=0; /*while(f[j-1]%prime[i]==0){ len++; f[j-1]/=prime[i]; } num[j-1]=(num[j-1]*((len*k+1)%mod))%mod;*/有毒的代码,换第一句RE,全换AC } } LL sum=0; for(int i=0;i<=r-l;i++){ if(f[i]!=1) num[i]=(num[i]*(k+1))%mod; sum=(sum+num[i])%mod; } printf("%lld\n",sum); } return 0;}
AC:
#include<bits/stdc++.h>using namespace std;typedef long long LL;#define mod 998244353#define maxn 1000001bool isprime[maxn];int cnt;LL prime[maxn],f[maxn],num[maxn],l,r,k;void Prime(){ cnt=0; memset(isprime,true,sizeof(isprime)); isprime[1]=false; for(int i=2;i<maxn;i++){ if(isprime[i]){ prime[cnt++]=i; for(int j=2;j*i<maxn;j++) isprime[i*j]=false; } }}int main(){ Prime(); int t; scanf("%d",&t); while(t--){ scanf("%lld%lld%lld",&l,&r,&k); for(int i=0;i<=r-l;i++) f[i]=i+l,num[i]=1; for(int i=0;i<cnt;i++){ LL st=(l/prime[i]+(l%prime[i]?1:0))*prime[i]; for(LL j=st;j<=r;j+=prime[i]){ LL len=0; while(f[j-l]%prime[i]==0) len++,f[j-l]/=prime[i]; num[j-l]=(num[j-l]*((len*k+1)%mod))%mod; } } LL sum=0; for(int i=0;i<=r-l;i++){ if(f[i]!=1) num[i]=(num[i]*(k+1))%mod; sum=(sum+num[i])%mod; } printf("%lld\n",sum); } return 0;}
阅读全文
0 0
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- POJ2478 Farey Sequence
- ----- UVA 11624-Fire!
- windows 如何定时更新svn
- Maven仓库镜像
- 二叉树的基本应用及其他
- 2017 Multi-University Training Contest
- Git中fetch和pull命令的区别
- sudo 命令能为你做些什么
- 求解最大连续子数组问题
- (2)window方法:confirm、setTimeout、setInterval、clearInterval、moveBy和moveTo、open和close
- CodeForces
- HDU 6070 Dirt Ratio 线段树 + 二分
- 201412-2 Z字形扫描
- 问题 E: YK的书架