LightOJ 1007(欧拉函数打表+前缀和)

Mathematically some problems look hard. But with the help of the computer, some problems can be easily solvable.

In this problem, you will be given two integers a and b. You have to find the summation of the scores of the numbers from a to b (inclusive). The score of a number is defined as the following function.

score (x) = n2, where n is the number of relatively prime numbers with x, which are smaller than x

For example,

For 6, the relatively prime numbers with 6 are 1 and 5. So, score (6) = 22 = 4.

For 8, the relatively prime numbers with 8 are 1, 3, 5 and 7. So, score (8) = 42 = 16.

Now you have to solve this task.
Input

Input starts with an integer T (≤ 105), denoting the number of test cases.

Each case will contain two integers a and b (2 ≤ a ≤ b ≤ 5 * 106).
Output

For each case, print the case number and the summation of all the scores from a to b.
Sample Input

Output for Sample Input

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6 6

8 8

2 20

Case 1: 4

Case 2: 16

Case 3: 1237
Note

Euler’s totient function applied to a positive integer n is defined to be the number of positive integers less than or equal to n that are relatively prime to n. is read “phi of n.”

Given the general prime factorization of , one can compute using the formula

题解：直接计算每个数的欧拉值然后平方和计算前缀和再相减就好了

#include<cstring>
#include<cstdio>
#include<iostream>
#include<string>
#include<queue>
#include<vector>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<stack>
#include<functional>
using namespace std;
#define eb(x) emplace_back(x)
#define pb(x) push_back(x)
#define ps(x) push(x)
#define clr(a,b) memset(a,b,sizeof(a))
#define _INIT ios::sync_with_stdio(false);cin.tie(nullptr);cout.precision(10); cout << fixed
#define lowbit(x) x&-x
#define inf 0x3f3f3f3f3f3f
#define MAX_N 200000+5
#define MAX_M 25
typedef long long ll;
typedef unsigned long long ull;
const ll INF=10000000000000000LL;
const ll maxn=5*1e6+2;
typedef priority_queue<int,vector<int>,less<int> > pql;
typedef priority_queue<int,vector<int>,greater<int> >pqg;
int prime[maxn];
ull eule[maxn];
void init()
{
    clr(eule,0);
    int tot=0;
    eule[1]=1;
    for(int i=2;i<maxn;i++)
    {
        if(!eule[i]){eule[i]=i-1;prime[tot]=i;tot++;}
        for(int j=0;j<tot&&i*prime[j]<maxn;j++)
        {
            eule[i*prime[j]]=1;
            if(i%prime[j]==0){eule[i*prime[j]]=eule[i]*prime[j];break;}//性质2和3
            else {eule[i*prime[j]]=eule[i]*eule[prime[j]];}//性质2
        }
    }
    for(int i=2;i<maxn;i++)
    {

        eule[i]=eule[i-1]+eule[i]*eule[i];
    }
    return ;
}
int main()
{
#ifndef ONLINE_JUDGE
    // freopen("data.txt","r",stdin);
#endif
    //_INIT;
    init();
    int t;
    scanf("%d",&t);
    for(int ti=1;ti<=t;ti++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("Case %d: %llu\n",ti,eule[b]-eule[a-1]);
    }
    return 0;
}

LightOJ 1007(欧拉函数打表+前缀和)

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