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HDU2710_Max Factor【水题】【筛法求素数】
Max Factor
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3966 Accepted Submission(s): 1289
Problem Description
To improve the organization of his farm, Farmer John labels each of his N (1 <= N <= 5,000) cows with a distinct serial number in the range 1..20,000. Unfortunately, he is unaware that the cows interpret some serial numbers as better than others. In particular, a cow whose serial number has the highest prime factor enjoys the highest social standing among all the other cows.
(Recall that a prime number is just a number that has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by 2 and 3, is not).
Given a set of N (1 <= N <= 5,000) serial numbers in the range 1..20,000, determine the one that has the largest prime factor.
Input
* Line 1: A single integer, N
* Lines 2..N+1: The serial numbers to be tested, one per line
Output
* Line 1: The integer with the largest prime factor. If there are more than one, output the one that appears earliest in the input file.
Sample Input
4
36
38
40
42
Sample Output
38
Source
USACO 2005 October Bronze
题目大意:给你N个数,求这N个数中哪个数的最大素因子最大,
输出这个数,如果有多种结果,输出靠前边的那个数。
思路:将筛法求素数改变一下。若i为素数,则i的1、2、3…倍的
最大素因子都为i,筛的时候,赋值为Prime[j] = i,即j的最大素因
子为i。
注意:初始化时令所有数为0,Prime[0] = Prime[1] = 1。
即Prime[i]为0是素数,Prime[i]为1为素数。改变之后Prime[i]为
i的最大素因子。
#include<stdio.h> #include<string.h> #include<math.h> int Prime[20005]; void IsPrime() { Prime[1] = 1; for(int i = 2; i <= 20000; i++) { if(Prime[i]==0) { Prime[i]=i; for(int j = i+i; j <= 20000; j+=i) Prime[j] = i; } } } int main() { int N,x; IsPrime(); while(~scanf("%d",&N)) { int Max = 0; while(N--) { scanf("%d",&x); if(Prime[x] > Prime[Max]) Max = x; } printf("%d\n",Max); } return 0; }
HDU2710_Max Factor【水题】【筛法求素数】
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