Problem Description
Euler is a well-known matematician, and, among many other things, he discovered that the formulan2 + 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 knownthat 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 certaininterval.InputEach line of input will be given two positive integer a and b such that 0 ≤ a ≤ b ≤ 10000. You mustread until the end of the file.OutputFor each pair a, b read, you must output the percentage of prime numbers produced by the formula inthis interval (a ≤ n ≤ b) rounded to two decimal digits.
Sample Input
0 39
0 40
39 40
Sample Output
100.00
97.565
0.00
题解：注意高精度，最后加个1e-8就行了。（不知道为啥->_->)
<span style="font-family:SimSun;">#include<cstdio> int s[10010]; int prime(int n) { for(int i=2;i*i<=n;i++) if(n%i==0) return 1; return 0; } int main() { for(int i=0;i<=10000;i++) s[i]=prime(i*i+i+41); double a,b,sum; while(~scanf("%lf%lf",&a,&b)) { sum=0; for(int i=a;i<=b;i++) { if(!s[i]) sum+=1; } double v; v=sum/(b-a+1)*100; printf("%.2lf\n",v+1e-8);//高精度，不加不对。 } return 0; }</span>
