HDOJ The Euler function 2824【欧拉函数】

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;
}