HDU 5317 RGCDQ (factor decomposition + pretreatment)

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Test instructions

The largest greatest common divisor of the species of the l,r of the range [of] all the number of elements ... Well, I don't know how to describe it.

For example, all the number of elements in the range of the element species are 1,2,3,4,5,6,7 then the result is gcd (3,6)

Analysis:

The range of data is 1~1000000, the maximum number of factors is 7, we first have to preprocess the number of all the number of factors, but because

The range of data is relatively large we cannot directly force the result because the number of factors may be smaller, so we can place

The prefix and the number of each value are calculated, and then the number of each value in the interval can be calculated within the time of O (1).

The code is as follows:

#include <iostream> #include <cstring> #include <cstdio> #include <algorithm>using namespace    std;const int maxn = 1e3+10;int pri[maxn],cnt;bool vis[maxn];void getprime () {cnt=0;    memset (vis,0,sizeof (VIS));            for (int i=2;i<maxn;i++) {if (!vis[i]) {pri[cnt++]=i;        for (int j=i+i;j<maxn;j+=i) vis[j]=1;    }}}int num[1000001];int sum[1000001][8];void init () {getprime ();    int ff=0;        for (int i=0;i<1000001;i++) {int tmp = i;        num[i]=0;                for (int j=0;j<cnt&&pri[j]*pri[j]<=tmp;j++) {if (tmp%pri[j]==0) {num[i]++;            while (tmp%pri[j]==0) tmp/=pri[j];        }} if (tmp>1) num[i]++;    Ff=max (NUM[I],FF);    } sum[0][0]=0;sum[0][1]=0;sum[0][2]=0;sum[0][3]=0;    sum[0][4]=0;sum[0][5]=0;sum[0][6]=0;sum[0][7]=0;        for (int. i=1;i<1000001;i++) {for (int j=1;j<=7;j++) SUM[I][J]=SUM[I-1][J]; Sum[i][num[i]]++;    }}int A[10];int Main () {init ();    int t,l,r;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &l,&r);        int tag = 0;        for (int i=7;i>=1;i--) a[i]=sum[r][i]-sum[l-1][i];                for (int i=7;i>=3;i--) {if (a[i]>=2) {tag = i;            Break        }} if (tag) {printf ("%d\n", tag);                } else{if (a[3]>=1&&a[6]>=1) {puts ("3");            Continue } else if ((a[2]>=1&&a[4]>=1) | | |                a[2]>=2) {puts ("2");            Continue        } else puts ("1"); }} return 0;}

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HDU 5317 RGCDQ (factor decomposition + pretreatment)