HDU 2710 Max Factorv (素数模板 & 多种解法)
Max Factor
http://acm.hdu.edu.cn/showproblem.php?pid=2710
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2412 Accepted Submission(s): 734
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.
(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
Source
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#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=20010; int prime[N]; int isprime[N]; void getPrime(){ //得到小于等于 N的素数，prime[0]存放的是个数 int i,j; prime[0]=0; for(i=0;i<N;i++) isprime[i]=1; for(i=2;i<N;i++) if(isprime[i]){ prime[++prime[0]]=i; for(j=2;i*j<N;j++) isprime[i*j]=0; } } long long factor[100][2]; int facCnt; int getFactors(long long x){ //把x进行素数分解 facCnt=0; long long tmp=x; for(int i=1;prime[i]*prime[i]<=tmp;i++){ factor[facCnt][1]=0; if(tmp%prime[i]==0){ factor[facCnt][0]=prime[i]; while(tmp%prime[i]==0){ factor[facCnt][1]++; tmp/=prime[i]; } facCnt++; } } if(tmp!=1){ factor[facCnt][0]=tmp; factor[facCnt++][1]=1; } return facCnt; } int main(){ //freopen("input.txt","r",stdin); getPrime(); int n; while(~scanf("%d",&n)){ int ans=1,tmp=1,num; for(int i=0;i<n;i++){ scanf("%d",&num); if(num==1) //1的时候要单独处理一下 continue; getFactors(num); if(tmp<factor[facCnt-1][0]){ tmp=factor[facCnt-1][0]; ans=num; } } printf("%d\n",ans); } return 0; }
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=20000+10; int prime[N]; int isprime[N]; void Init(){ int i,j; int cnt=0; for(i=0;i<N;i++) isprime[i]=1; for(i=2;i<N;i++) if(isprime[i]){ prime[cnt++]=i; for(j=2;i*j<N;j++) isprime[i*j]=0; } } int main(){ //freopen("input.txt","r",stdin); Init(); int n; while(~scanf("%d",&n)){ int num,ans=1,max=1,tmp; while(n--){ scanf("%d",&num); for(int i=0;prime[i]<=num;i++) if(num%prime[i]==0) tmp=prime[i]; if(tmp>max){ max=tmp; ans=num; } } printf("%d\n",ans); } return 0; }
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int prime[20010]; int isPrime(int x){ for(int i=2;i*i<=x;i++) if(x%i==0) return 0; return 1; } int main(){ //freopen("input.txt","r",stdin); memset(prime,0,sizeof(prime)); prime[1]=1; for(int i=2;i<=20000;i++) if(isPrime(i)) prime[i]=1; int n; while(~scanf("%d",&n)){ int ans=1,tmp=1,max=1,res=1; int num; while(n--){ scanf("%d",&num); int j=1; while(1){ if(num%j==0 && prime[num/j]){ tmp=num/j; res=num; break; } j++; } if(tmp>max){ max=tmp; ans=num; } } printf("%d\n",ans); } return 0; }
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int prime[20010]; void Init(){ memset(prime,0,sizeof(prime)); for(int i=2;i<=20000;i++) if(!prime[i]) for(int j=i*i;j<=20000;j+=i) prime[j]=1; } int main(){ //freopen("input.txt","r",stdin); Init(); int n; while(~scanf("%d",&n)){ int ans=1,tmp=1,max=1,res=1; int num; while(n--){ scanf("%d",&num); int j=1; while(1){ if(num%j==0 && !prime[num/j]){ tmp=num/j; res=num; break; } j++; } if(tmp>max){ max=tmp; ans=num; } } printf("%d\n",ans); } return 0; }
下面这题，哎，，被蛋疼的 2 WA了两个小时：http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2093
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=1000000; int prime[N+10],isprime[N+10]; int cnt,num[N+10]; void getPrime(){ int i,j; prime[0]=0; for(i=0;i<N;i++) isprime[i]=1; for(i=2;i<N;i++) if(isprime[i]){ prime[++prime[0]]=i; for(j=2;i*j<N;j++) isprime[i*j]=0; } } void init(){ getPrime(); cnt=0; for(int i=1;i<=prime[0];i++) if((prime[i]&1) && prime[i]%4==1) num[cnt++]=prime[i]; } int main(){ //freopen("input.txt","r",stdin); init(); int L,U; while(~scanf("%d%d",&L,&U)){ if(L==-1 && U==-1) break; int a=0,b=0; int l=L,u=U; if(L<0) L=0; if(U<0) U=0; if(L<=2 && U>=2) //2=1*1+1*1 你懂得 b++; for(int i=1;i<=prime[0];i++) if(prime[i]>=L && prime[i]<=U) a++; for(int i=0;i<cnt;i++) if(num[i]>=L && num[i]<=U) b++; printf("%d %d %d %d\n",l,u,a,b); } return 0; }
