1013 数素数

solution1

直接判断

#include<stdio.h>
#include<math.h>
int main(){
	int n, m, count = 0, flag, c = 0;
	scanf("%d%d", &m, &n);
	for(int i = 2; count < n; i++){
		flag = 0;
		for(int j = 2; j <= sqrt(i * 1.0); j++){
			if(i % j == 0){
				flag = 1; 
				break;
			}
		}
		if(!flag){
			count++;
			if(count >= m){
				c++;
				printf("%d", i);
				if(c % 10 != 0 && c != n - m + 1) printf(" ");
				if(c % 10 == 0) printf("\n");
			}
		}
	}
	return 0;
}

solution2 埃氏筛法

#include<stdio.h>
const int maxN = 1000001;
int main(){
	int hash[maxN] = {0}, p[maxN], pos = 0, n, m, count = 0;
	scanf("%d%d", &m, &n);
	for(int i = 2; i < maxN; i++){
		if(!hash[i]){
			p[pos++] = i;
			if(pos >= n) break;
			for(int j = 2 * i; j < maxN; j += i) hash[j] = 1;
		}
	}
	for(int i = m - 1; i < n; i++){
		count++;
		printf("%d", p[i]);
		if(count % 10 != 0 && count != n - m + 1) printf(" ");
		if(count % 10 == 0) printf("\n");
	}
	return 0;
}