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
#include<cstdio> #include<cstring> int pa[10011]={1,1}; int node(int a) { int i; for(i=2;i*i<=a;i++) { if(a%i==0) return 0; } return 1; } int main() { int i; memset(pa,0,sizeof(pa)); for(i=0;i<=10000;i++) pa[i]=node(i*i+i+41);//打表 int a,b; int ans; while(scanf("%d%d",&a,&b)!=EOF) { ans=0; for(i=a;i<=b;i++) ans+=pa[i]; printf("%.2lf\n",ans*1.0/(b-a+1)*100+1e-8);//1e-8精度 } return 0; }