3 条题解
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1房佳坤 LV 10 @ 2021-11-22 20:26:46
#include <bits/stdc++.h> using namespace std; const int N = 100010; int primes[N], cnt; bool st[N]; void get_primes(int m) // 线性筛 { for ( int i = 2; i <= m; i ++ ) { if ( !st[i] ) primes[cnt ++] = i; for ( int j = 0; primes[j] <= m / i; j ++ ) { st[primes[j] * i] = 1; if ( !( i % primes[j] ) ) break; } } } int main() { int n, m; cin >> n >> m; get_primes(m); int res = 0; for ( int i = 0; i < cnt; i ++ ) if ( primes[i] >= n && primes[i] <= m ) res ++ ; cout << res; return 0; }
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02019-04-20 16:08:17@
a=int(input()) b=int(input()) d=0 def prime(x): for i in range(2,x): if x%i==0: return(0) return(1) for g in range(a,b+1): if prime(g)==1 and g>=2: d=d+1 print(d)
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02019-04-19 15:01:33@
import math a=int(input()) b=int(input()) cnt=0 for i in range(a,b+1): flag = 1 if i==1: continue if (i == 2): cnt+=1 a1 = int(math.ceil(math.sqrt(i)) + 1) for j in range(2, a1): if i % j == 0: flag=0 if flag: cnt+=1 print(cnt)
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- 难度
- 7
- 分类
- (无)
- 标签
- (无)
- 递交数
- 638
- 已通过
- 121
- 通过率
- 19%
- 被复制
- 3
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