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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 formulaoutput 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. SampleInput
0 39
0 40
39 40
Sample Output
100.00
97.56
50.00
#include<stdio.h> #include<math.h> #include<algorithm> #include<string.h> using namespace std; #define G 11010 int num[G]; int sushu(int n) { int k = sqrt(n); //这里用n/2超时 for(int i = 2 ; i <= k ; i++) { if(n % i == 0) return 0; } return 1; } void init() { num[0]=1; for(int i = 1 ; i <G ; i++) num[i] = num[i - 1] + sushu(i * i + i + 41); } int main() { int a, b,sum; double p; init(); while(scanf("%d%d", &a, &b)!=EOF) { sum= num[b] - num[a] + sushu(a * a + a + 41); //printf("***%d****\n",sum); p=(double)sum/(b-a+1)*100+ 1e-8;//这里如果不提高精度，会WA； printf("%.2f\n",p); } return 0; }
