#问题来源于HDOJ2136
#原题:
Largest prime factor
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15784 Accepted Submission(s): 5486
Problem Description
Everybody knows any number can be combined by the prime number.
Now, your task is telling me what position of the largest prime factor.
The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc.
Specially, LPF(1) = 0.
Now, your task is telling me what position of the largest prime factor.
The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc.
Specially, LPF(1) = 0.
Input
Each line will contain one integer n(0 < n < 1000000).
Output
Output the LPF(n).
Sample Input
12345
Sample Output
01213
#AC码:
//#include<iostream> #include<stdio.h> #include<string.h> //using namespace std; #define max 1000000 int num[max]; void perp() { int cn=1; for(int i=2;i<max;i++) if(num[i]==0) { for(int j=1;j*i<max;j++) num[j*i]=cn; cn++; } } int main() { memset(num,0,sizeof(num)); int n; perp(); while(scanf("%d",&n)!=EOF) printf("%d\n",num[n]); }
#问题:如果将scanf和printf换成cin和cout,则Time Limit Exceeded
#原因分析:(查找到相关博客点击打开链接)
仅仅IO速度差别!
以后在大量IO时一定要注意这点。
本文介绍了HDOJ2136问题——寻找整数的最大质因数的位置,并提供了一种有效算法实现。通过预先计算所有小于100万的数的最大质因数位置,实现了快速查询,特别指出在大量输入输出操作中使用scanf和printf比cin和cout更高效。

被折叠的 条评论
为什么被折叠?



