Mathematically Hard LightOJ

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
3
6 6
8 8
2 20
Sample Output
Case 1: 4
Case 2: 16
Case 3: 1237

大致题意：就是让你求区间[a,b]内的各个数的欧拉值平方的和

思路：先用欧拉筛打个表，然后用前缀和预处理下，注意最后结果会爆longlong，得用unsign long long，然后内存也得注意下，写搓了很容易爆。

代码如下

#include<bits/stdc++.h>using namespace std;#define LL long long #define ULL unsigned long long const int M=5000010;ULL p[M];//先用于欧拉筛里存素数，然后再用来保存前缀和，如果再开一个数组的话会爆内存int phi[M];bool vis[M];int p_num=0;void get_phi(){    for(int i=2;i<=M;i++)    {        if(!vis[i])        {            p[p_num++]=i;            phi[i]=i-1;        }        for(int j=0;j<p_num&&p[j]*i<=M;j++)        {            vis[p[j]*i]=1;            if(i%p[j]==0)            {                phi[i*p[j]]=phi[i]*p[j];                break;            }            else             phi[i*p[j]]=phi[i]*(p[j]-1);        }    }}int main(){    get_phi();    phi[1]=1;    p[0]=0;    for(int i=1;i<=M;i++)    {        p[i]=(ULL)phi[i]*phi[i]+p[i-1];//注意phi[i]*phi[i]会爆int    }    int T,a,b;    scanf("%d",&T);    for(int cas=1;cas<=T;cas++)    {        scanf("%d%d",&a,&b);        printf("Case %d: %llu\n",cas,p[b]-p[a-1]);    }}