HDU 2012 素数判定
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 72727 Accepted Submission(s): 25323
Problem Description
对于表达式n^2+n+41,当n在(x,y)范围内取整数值时(包括x,y)(-39<=x<y<=50),判定该表达式的值是否都为素数。
Input
输入数据有多组,每组占一行,由两个整数x,y组成,当x=0,y=0时,表示输入结束,该行不做处理。
Output
对于每个给定范围内的取值,如果表达式的值都为素数,则输出"OK",否则请输出“Sorry”,每组输出占一行。
Sample Input
0 1 0 0
Sample Output
OK
#include<iostream> using namespace std; bool bf[2600]; void getss(){ for(int i=2;i<2600;i++) { if(!bf[i]){ for(int j=2;j*i<2600;j++){ bf[i*j]=true; } } } } int main(){ memset(bf,false,sizeof(bf)); getss(); int x,y; bool is; while(cin>>x>>y){ if(x==0&&y==0)break; is=true; for(int k=x;k<=y;k++){ if(bf[k*k+k+41]==true){ is=false;break; } } if(is==true) cout<<"OK"<<endl; else cout<<"Sorry"<<endl; } return 0; }
FZU Problem 1756 FactorSum
Accept: 429 Submit: 1134
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Give you a signed integer n (n<=10000), please tell me the sum of its all non-negative factors, and all its factors must larger than zero. Say, give you 12, you know, 1,2,3,4,6 are its factors. So the answer would be 1+2+3+4+6=16. if n < 0 , just output "0"
There are several test cases, for evey case, there is exactly one line containing an integer N (N<=10000).
Sample Input
1234
Sample Output
0113
思路:对于每个数字,他的倍数都要加上他;比如 我们一共要求到6;那么从一开始是一的倍数(不超过6)的都要加1;直到6;然后在从2开始 以此类推;
#include<iostream> using namespace std; int bf[10001]; int main(){ memset(bf,0,sizeof(bf)); for(int i=1;i<=10000;i++) for(int j=2;i*j<=10000;j++){ bf[i*j]+=i; } int n; while(cin>>n){ if(n<0)cout<<0<<endl; else cout<<bf[n]<<endl; } return 0; }
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