Time limit: 20000ms single point time limit: 1000ms memory limit: 256MB description
Fragrance today is in a good mood, and the flowers in the field of insects play a puzzle game.
This game is like this, for an array A, fragrance Select a number a from a, insects select a number b from a. A and B can be the same. Their score is the number of a*b factors.
The delicate fragrance and the worm of course want to get as high a score as possible, can you tell them which two number to choose?
Since the fragrance is a very casual person, the elements in array A are randomly chosen by her.
Input
A row of number N, which represents the number of integers in a.
The next line of N is the number of A1,A2,..., an, respectively, as an element of a.
n <= 100000, 1 <= ai <= 100000
Ensure that all AI are randomly generated.
Output
A row represents the maximum score.
How to think there is no algorithm, in fact, the right violent posture can also be a miracle.
At first, we only think of the number of factors b[i], and then skillfully use break;
This form: if (Ans>b[i]*b[j]) break a layer of loops, b arrays are sorted by the number of factors from large to small, so correct.
But each time we calculate the sum of the number of factors (a[i],a[j]), we have to enumerate through the past, TLE.
For example, this code:
#include <cstdio> #include <algorithm> #include <cmath> #include <string.h> #include <string >using namespace std; #define N 100001int p[n];int b[n],t;struct node{int x, y;} A[n];void Prime () {for (Int. i=2;i<n;i++) if (!b[i]) for (int j=i+i;j<n;j+=i) b[j]=1; for (int i=2;i<n;i++) if (!b[i]) p[++t]=i;} int work (int x,int y) {int sum=1; int t=1; while (1) {if (x==1&&y==1) return sum; int v=1; if (x%p[t]==0) {while (x%p[t]==0) {v++; X/=P[T]; }} if (Y%p[t]==0) {while (y%p[t]==0) {v++; Y/=P[T]; }} t++; Sum*=v; }}int CMP (node A,node b) {return a.y>b.y;} int main () {prime (); Memset (b,0,sizeof (b)); int n; scanf ("%d", &n); for (int i=1;i<=n;i++) scanf ("%d", &a[i].x), A[i].y=work (a[i].x,1); for (int i=1;i<=n;i++) printf ("%d", a[i].y); Sort (a+1,a+n+1,cmp); int ans=0; for (int i=1;i<=n;i++) for (int. j=i+1;j<=n;j++) {if (ans>a[i].y*a[j].y) break; else Ans=work (a[i].x,a[j].x); } printf ("%d\n", ans); return 0;}
So I see a template: Linear handsome, and each time we can find a number of the precursor prime factor
1#include <cstdio>2#include <algorithm>3#include <cmath>4#include <string.h>5#include <string>6 7 using namespacestd;8 9 #defineN 100001Ten One intP[n]; A intb[n],t=0; - - structnode the { - intx, y; - }a[n]; - voidPrime () + { - for(intI=2; i<n;i++) + { A if(!b[i]) p[++t]=b[i]=i; at for(intj=1; j<=t&&p[j]<=b[i]&&p[j]<=n/i;++j) b[i*p[j]]=P[j]; - } -b[1]=N; - } - - intWorkintXinty) in { - intsum=1; to intC=0; + while(x>1|| Y>1) - { the if(B[x]<b[y]) c=B[x]; * ElseC=B[y]; $ intv=1;Panax Notoginseng for(; b[x]==c;x/=c) v++; - for(; b[y]==c;y/=c) v++; thesum*=v; + } A returnsum; the } +InlineintCMP (node A,node b) - { $ returnA.y>b.y; $ } - - intMain () the { - Prime ();Wuyi the intN; -scanf"%d",&n); Wu for(intI=1; i<=n;i++) -scanf"%d", &a[i].x), A[i].y=work (a[i].x,1); About $Sort (A +1, a+n+1, CMP); - intans=a[1].y; -N=min (N, $); - A for(intI=1; i<=n;i++) + for(intj=i+1; j<=n;j++){ the if(ANS>A[I].Y*A[J].Y) Break; - Elseans=Max (Ans,work (a[i].x,a[j].x)); $ } the theprintf"%d\n", ans); the return 0; the}
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But you will find still very slow, super slow, 9000ms, and then look at a code, found!!
We enumerate the
for (int i=1;i<=n;i++) for (int j=i+1;j<=n;j++)
Because n can be very large, about 100,000, but the back is mostly useless, so we just n<=1000, and so on is enough.
Violence is horrible.
Hihocoder 1165: Puzzle game