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.
Input
Each 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
97.56
50.00
给出一个范围a->b，然后求满足关系的素数率.需要素数打表
代码如下：
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int sum[10010]; bool prime(int x) { int i; for(i=2;i*i<=x;i++) { if(x%i==0) return false; } return true; } void getprime() { sum[0]=1; for(int i=1;i<10010;i++) { if(prime(i*i+i+41)) sum[i]=sum[i-1]+1; else sum[i]=sum[i-1]; } } int main() { getprime(); int a,b; while(scanf("%d%d",&a,&b)!=EOF) { int ans=sum[b]-sum[a]+prime(a*a+a+41);//求满足条件的个数. double k=1.0*ans/(b-a+1); k=(int)(k*10000+0.5)/100.0;//四舍五入，100后面需要加.0(刚开始一直错就是因为这个)； printf("%.2lf\n",k); } }
