HDU 5317 RGCDQ (prime factor decomposition + preprocessing)

HDU 5317 RGCDQ (prime factor decomposition + preprocessing)

 

Evaluate the maximum common divisor of all prime factor types in the range [l, r... Yes, I don't know how to describe it...

For example, if the number of all prime factors in the interval is 1, 2, 3, 4, 5, 6, and 7, the result is gcd (3, 6)

Analysis:

The data range is 1 ~ 1000000, the maximum number of factors is 7. First, we need to preprocess the number of factors in all numbers, but because

If the data range is large, we cannot directly obtain the results. because the number of factors may have a small value, We Can prefix

Calculate the prefix and number of each value, and then you can find the number of each value in the interval in the time of O (1.

The Code is as follows:

#include 
 
  #include 
  
   #include 
   
    #include 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
    
     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,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;}