Description
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.
0 39
0 40
39 40
100.00
97.56
50.00
输出结果要加上1e-8就是 0.00000001 保证高精度；
代码：
#include<stdio.h> #include<string.h> int su[10002]; bool f(int t) { for(int i=2;i*i<=t;i++) { if(t%i==0) { return 0; } } return 1; } int main() { for(int i=0;i<=10002;i++) { su[i]=f(i*i+i+41); } int a,b; while(scanf("%d%d",&a,&b)!=EOF) { int i=0; int ans=0; for(i=a;i<=b;i++) { //if(su[i]) //ans++; ans+=su[i]; } double p=0; p=ans*1.0/(b-a+1)*100; printf("%.2lf\n",p+1e-8);//高精度，不加1e-8wa到死；；；；；；；；；； } return 0; }
