题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1141
Prime Time
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
那么水的为什么比赛时没做出来呢，在想啥呢。 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int init(int n) { for(int i=2;i*i<=n;i++) if(n%i==0) { return 0; } return 1; } int main() { int a,b,i,j,num[10010]; memset(num,0,sizeof(num)); for(i=0;i<10010;i++) num[i]=init(i*i+i+41); int sum; while(scanf("%d%d",&a,&b)!=EOF) { sum=0; for(i=a;i<=b;i++) sum+=num[i]; printf("%.2lf\n",sum*1.0/(b-a+1)*100+1e-8); } return 0; }
