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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 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 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 |
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; }
Lightoj 1007 - Mathematically Hard
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原文地址：http://www.cnblogs.com/pk28/p/5574137.html
