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.
Sample Input
0 39
0 40
39 40
Sample Output
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;
}