Lightoj 1007 - Mathematically Hard

1007 - Mathematically Hard

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Time Limit: 2 second(s)

Memory Limit: 64 MB

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
3 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

 

欧拉函数打表。抄的大白书上的板子。一开始没看note  自己yy筛法。果然很弱智，题目已经给了Note了，眼瞎。

/* ***********************************************
Author        :guanjun
Created Time  :2016/6/10 19:35:28
File Name     :1007.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 5000100
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int phi[maxn];
ull sum[maxn];
void phi_table(int n){
    for(int i=2;i<=n;i++)phi[i]=0;
    phi[1]=1;
    for(int i=2;i<=n;i++)if(!phi[i]){
        for(int j=i;j<=n;j+=i){
            if(!phi[j])phi[j]=j;
            phi[j]=phi[j]/i*(i-1);
        }
    }
    sum[1]=0;
    for(int i=2;i<=maxn;i++){
        sum[i]=sum[i-1]+(ull)phi[i]*phi[i];
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    phi_table(maxn);
    int T,a,b;
    cin>>T;
    for(int t=1;t<=T;t++){
        scanf("%d%d",&a,&b);
        printf("Case %d: %llu\n",t,sum[b]-sum[a-1]);
    }
    return 0;
}