LightOJ 1007 - Mathematically Hard （欧拉筛+预处理前缀和）

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 ofthe computer, some problems can be easily solvable.

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

score (x) = n2, where n is thenumber 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 and7. 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 allthe 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 numberof 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 cancompute 

 using
the formula



题意：给你L和R，求这个范围内的每个数的欧拉函数的平方和

思路：看一下数据范围5e6，不大，打表就行，先筛一下欧拉数，然后预处理前缀和，因为最后结果会爆long long，所以用unsign long long

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 5000010
#define LL unsigned long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-10
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
LL ans[MAXN];
void db()
{
mem(ans);
ans[1]=1;
int i,j;
for(i=2;i<=MAXN;i++)
{
if(!ans[i])
{
for(j=i;j<=MAXN;j+=i)
{
if(!ans[j])
ans[j]=j;
ans[j]=ans[j]/i*(i-1);
}
}
}
for(i=1;i<=MAXN;i++)
ans[i]=ans[i-1]+ans[i]*ans[i];
}
int main()
{
db();
int t,l,r;
int cas=0;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&l,&r);
printf("Case %d: ",++cas);
printf("%llu\n",ans[r]-ans[l-1]);
}
return 0;
}