Spoj-ODDDIV Odd Numbers of Divisors

Given a positive odd integer K and two positive integers low and high, determine how many integers between low and high contain exactly K divisors.

Input

The first line of the input contains a positive integer C (0<C<100,000), the number of test cases to follow. Each case consists of a line containing three integers: K, low, and high (1<K<10000, 0<low≤ high<10^10). K will always be an odd integer.

Output

Output for each case consists of one line: the number of integers between low and high, inclusive, that contain exactly K divisors.

Example

Input:
3
3 2 49
9 1 100
5 55 235

Output:
4
2
1

询问一组(k,l,r)，意思是在数字l到r之间有多少个数字有奇数个因子

显然如果对一个x质因数分解成(大π) pi^qi  那么总因子数是(大π) (qi+1)

因为它有奇数个因数，所以所有qi都是偶数，所以x应当是完全平方数！

因此，只要枚举x，而且l和r的范围从1e10将到1e5

把一个询问(k,l,r)差分成(k,1,r)-(k,1,l-1),就可以离线搞了

然后直接从1到10w枚举每个x，计算x^2有多少个因子，更新对于一个k，当前1~k已经记录了多少个完全平方数

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 inline void write(LL a)
 {
     if (a<){printf("-");a=-a;}
     if (a>=)write(a/);
     putchar(a%+'');
 }
 inline void writeln(LL a){write(a);printf("\n");}
 LL l,r,k,n;
 int mx;
 bool isprime[];
 int sum[];
 struct ask{int k,n,rnk;}q[];
 bool operator <(ask a,ask b){return a.n<b.n;}
 int cnt[];
 int ans[];
 inline void init()
 {
     memset(isprime,,sizeof(isprime));
     for (int i=;i<=;i++)sum[i]=;
     isprime[]=isprime[]=;
     int k,l;
     for (int i=;i<=;i++)
     {
         if (isprime[i])
         {
             sum[i]=;
             for (int j=;i*j<=;j++)
             {
                 isprime[i*j]=;k=;l=j;
                 while (l%i==){l/=i;k++;}
                 sum[i*j]*=*k+;
             }
         }
     }
 }
 int main()
 {
     init();
     n=read();
     for (int i=;i<=n;i++)
     {
         int x=read(),y=ceil(sqrt(read())),z=floor(sqrt(read()));
         mx=max(mx,z);
         q[*i-].k=x;q[*i-].n=y-;q[*i-].rnk=-i;
         q[*i].k=x;q[*i].n=z;q[*i].rnk=i;
     }
     sort(q+,q+*n+);
     int now=;
     for (int i=;i<=mx;i++)
     {
         cnt[sum[i]]++;
         while (now<=*n&&q[now].n==i)
         {
             if (q[now].rnk>)ans[q[now].rnk]+=cnt[q[now].k];
             else ans[-q[now].rnk]-=cnt[q[now].k];
             now++;
         }
     }
     for (int i=;i<=n;i++)printf("%d\n",ans[i]);
 }

Spoj ODDDIV