HDU 2710 Max Factor 数论 (水
Max Factor
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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.
(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
* 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
比较水xjb做
新风格不错 以后就这样写了
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <queue> #include <vector> #include <iomanip> #include <math.h> #include <map> using namespace std; #define FIN freopen("input.txt","r",stdin); #define INF 0x3f3f3f3f #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef long long LL; const int MAXN=20005; int p[MAXN]; int main() { //FIN p[1]=1; int cou=0; for(int i=2; i<=20005; i++) { if(p[i]==0) { cou++; for(int j=i; j<=20005; j=j+i) { p[j]=cou; } } } int n; while(~scanf("%d",&n)) { int ans=0,pos; int x; for(int i=1;i<=n;i++){ scanf("%d",&x); if(p[x]>ans) {ans=p[x];pos=x;} } printf("%d\n",pos); } }