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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 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).
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 |
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; }