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HDOJ The Euler function 2824【欧拉函数】

2015-07-31 17:29 218 查看

The Euler function

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4389 Accepted Submission(s): 1821



Problem Description

The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose
you are given a, b, try to calculate (a)+ (a+1)+....+ (b)


Input

There are several test cases. Each line has two integers a, b (2<a<b<3000000).


Output

Output the result of (a)+ (a+1)+....+ (b)


Sample Input

3 100




Sample Output

3042




Source

2009 Multi-University Training Contest 1 - Host
by TJU


Recommend

gaojie | We have carefully selected several similar problems for you: 2818 2825 2822 2821 2820

打表求欧拉函数值

注意数组开成long long 型

第一种：稍微快一点

#include <stdio.h>
#include <math.h>
#include <vector>
#include <queue>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 3000000+100

using namespace std;

long long E[maxn];

void euler()
{
    E[1]=1;
    for(int i=2;i<maxn;i++)
        E[i]=i;
    for(int i=2;i<maxn;i++){
        if(E[i]==i)
        for(int j=i;j<maxn;j+=i){
            E[j]=E[j]/i*(i-1);
        }
    }
    for(int i=2;i<maxn;i++)
        E[i]+=E[i-1];
}
int main()
{
    euler();
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF)
        printf("%lld\n",E[b]-E[a-1]);
    return 0;
}


第二种

#include <stdio.h>
#include <math.h>
#include <vector>
#include <queue>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 3000000+100

using namespace std;

long long E[maxn];

void euler()
{
    for(int i=2;i<maxn;i++){
        if(!E[i])
        for(int j=i;j<maxn;j+=i){
            if(!E[j])E[j]=j;
            E[j]=E[j]/i*(i-1);
        }
        E[i]+=E[i-1];
    }
}

int main()
{
    euler();
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF)
        printf("%lld\n",E[b]-E[a-1]);
    return 0;
}
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