点击打开题目
Euler is a well-known matematician, and, among many other things, he discovered that the formula n 2 + n + 41 produces a prime for 0 ≤ n < 40. For n = 40, the formula produces 1681, which is 41 ∗ 41. Even though this formula doesn’t always produce a prime, it still produces a lot of primes. It’s known that for n ≤ 10000000, there are 47,5% of primes produced by the formula! So, you’ll write a program that will output how many primes does the formula output for a certain interval.
Input Each line of input will be given two positive integer a and b such that 0 ≤ a ≤ b ≤ 10000. You must read until the end of the file.
Output For each pair a, b read, you must output the percentage of prime numbers produced by the formula in this interval (a ≤ n ≤ b) rounded to two decimal digits.
Sample Input 0 39 0 40 39 40
Sample Output 100.00 97.56 50.00
题意就是说用 n^2 + n +41 这个公式出来的数是素数的概率。
对0~10000打一个表就行了。
输出的时候注意四舍五入（这一点特别坑）
代码如下：
#include <stdio.h> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define CLR(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define LL long long #define f(x) (x*x+x+41) int sum[10000+11]; bool isPrime(int x) { int endd = sqrt(x); for (int i = 2 ; i <= endd ; i++) { if (x % i == 0) return false; } return true; } void getPrime() { sum[0] = 1; for (int i = 1 ; i <= 10000 ; i++) { if (isPrime(f(i))) sum[i] = sum[i-1] + 1; else sum[i] = sum[i-1]; } } int main() { getPrime(); int n; int l,r; double ans; while (scanf ("%d %d",&l,&r) != EOF) { if (l == 0) ans = (double)(sum[r]) / (r+1); else ans = (double)(sum[r] - sum[l-1]) / (r - l + 1); ans = (int)(ans*10000+0.5) / 100.0; printf ("%.2lf\n",ans); } return 0; }
