用for循环求100以内的素数python
按照你的要求编写的Python程序如下(输出每5个一行)
import math
count=1
for i in range(2,100):
for j in range(2,int(math.sqrt(i))+1):
if i%j==0:
break
else:
if count%5==0:
print(i)
else:
print(str(i)+"",end='')
count+=1
源代码(注意源代码的缩进)
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绛旓細for(n=2;n<=100;n++) // 鐢ㄥ彉閲 n 閬嶅巻 2~100锛屾壘鍏朵腑绱犳暟 { for(i=2;n%i;i=i++); // 璇曞浘鎵惧嚭鏈灏忕殑鍙橀噺 i 鍊硷紝浣垮緱 n%i == 0锛屼篃灏辨槸 n 鑳借 i 鏁撮櫎锛岀劧鍚庣粓姝㈣繖涓 for 寰幆銆俰f(i==n) // 鍒ゆ柇閫氳繃涓婇潰鐨 for 寰幆鎵惧埌鐨 i 鍊笺傚鏋 i 绛変簬 n 鏈韩锛岄偅涔 n 灏...
绛旓細include<stdio.h> include <math.h> void main(){ int i,j,sum;sum = 0;for(i=2; i<100; i++){ for(j = 2; j<=sqrt(i); j++){ if(i % j == 0){ break;} } if(j>sqrt(i)){ printf("%d ",i);sum += i;} } printf("\n");printf("绱犳暟鍜屼负锛%d\n",sum)...
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