1 solutions
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0
C :
#include<stdio.h> #include<math.h> int main() { int i,j,s=0; for(i=100;i<=200;i++) { for(j=2;j<=sqrt(i);j++) { if(i%j==0)break; } if(j>sqrt(i)){s++;if(s==1)printf("%d",i);else printf(" %d",i);}; } return 0; }C++ :
#include <iostream> #include<cmath> //在Dev C++中可调用数学函数库cmath using namespace std; int main () { int x; for (int i=100;i<=200;++i) { x=2; while(x<=floor(sqrt(i))&&(i%x!=0)) //floor为取整函数,需调用math.h库 x=x+1; //在枚举的范围内并且没有出现约数则继续枚举 if ( x>floor(sqrt(i))) cout<<i<<" "; } return 0; }
- 1
Information
- ID
- 16341
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
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- 0
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