B - RGCDQ-HDU 5317-素数筛法

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题意 f(x)是x能分解成的素数的种类，

http://acm.hdu.edu.cn/showproblem.php?pid=5317

先打素数表，再用素数筛选出每个数的种类

接下来求任意区间的元素对应的f(x)的最大公约数，----队友读漏了gcd，以为是求区间最大值，拍了个线段树-无用功....

因为答案显然只有1 2 3 4 5 6 7...所以预处理记下左端到每个数字间有多少个1 2 3 4 5 6 7，即可，然后对区间[a,b]记录的1 2 3 4 5 6 7个数相减，即可得到区间之间有多少个1 2 3 4 5，然后判断一下就好

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <iostream>#include <queue>#include <set>#include <vector>#define MAX 1000005#define inf 0x3f3f3f3fusing namespace std;#define tree_size MAX*3int max(int a,int b){if (a<b)return b;return a;};bool f[1000500]; int ans[1000500];  int yue[1000500][8];  int main(){//素数打表  __int64 i,j;f[1]=true;for (i=2;i<=MAX;i++)    //1处{if (f[i]==false)           //优化{for (j=(__int64)i*i;j<=1000500;j=j+i)   // 如果1处用1000必须_int64强制转换{f[j]=true;}}} for (i=2;i<=MAX;i++)    //1处{if (f[i]==false){for (j=i;j<=MAX;j=j+i){ans[j]++;} } }for (i=1;i<=1000000;i++){for (j=1;j<=7;j++)yue[i][j]=yue[i-1][j];yue[i][ans[i]]++;}int l,t,a,b;scanf("%d",&t);    while(t--)    {scanf("%d%d",&a,&b);int flag=0;for (j=7;j>=3;j--){int tmp=yue[b][j]-yue[a-1][j];if (tmp>=2){printf("%d\n",j);flag=1;break;}}if (flag) continue;int t1=yue[b][6]-yue[a-1][6];int t2=yue[b][3]-yue[a-1][3];if (t1+t2>=2)  {printf("3\n"); continue;}else{t1=yue[b][2]-yue[a-1][2];t2=yue[b][4]-yue[a-1][4];int t3=yue[b][6]-yue[a-1][6];if (t1+t2+t3>=2){ printf("2\n");continue; }elseprintf("1\n");} }return 0;}

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <iostream>#include <queue>#include <set>#include <vector>#define MAX 1000005#define inf 0x3f3f3f3fusing namespace std;#define tree_size MAX*3int max(int a,int b){if (a<b)return b;return a;};bool f[1000500]; int ans[1000500];  int yue[1000500][8];  int main(){//素数打表  __int64 i,j;f[1]=true;for (i=2;i<=MAX;i++)    //1处{if (f[i]==false)           //优化{for (j=(__int64)i*i;j<=1000500;j=j+i)   // 如果1处用1000必须_int64强制转换{f[j]=true;}}} for (i=2;i<=MAX;i++)    //1处{if (f[i]==false){for (j=i;j<=MAX;j=j+i){ans[j]++;} } }for (i=1;i<=1000000;i++){for (j=1;j<=7;j++)yue[i][j]=yue[i-1][j];yue[i][ans[i]]++;}int l,t,a,b;scanf("%d",&t);    while(t--)    {scanf("%d%d",&a,&b);int flag=0;for (j=7;j>=3;j--){int tmp=yue[b][j]-yue[a-1][j];if (tmp>=2){printf("%d\n",j);flag=1;break;}}if (flag) continue;int t1=yue[b][6]-yue[a-1][6];int t2=yue[b][3]-yue[a-1][3];if (t1+t2>=2)  {printf("3\n"); continue;}else{t1=yue[b][2]-yue[a-1][2];t2=yue[b][4]-yue[a-1][4];int t3=yue[b][6]-yue[a-1][6];if (t1+t2+t3>=2){ printf("2\n");continue; }elseprintf("1\n");} }return 0;}

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