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
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<deque> 9 #include<set> 10 #include<map> 11 #include<ctime> 12 #define LL long long 13 #define inf 0x7ffffff 14 #define pa pair<int,int> 15 #define pi 3.1415926535897932384626433832795028841971 16 using namespace std; 17 inline LL read() 18 { 19 LL x=0,f=1;char ch=getchar(); 20 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 21 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 22 return x*f; 23 } 24 inline void write(LL a) 25 { 26 if (a<0){printf("-");a=-a;} 27 if (a>=10)write(a/10); 28 putchar(a%10+'0'); 29 } 30 inline void writeln(LL a){write(a);printf(" ");} 31 LL l,r,k,n; 32 int mx; 33 bool isprime[100010]; 34 int sum[100100]; 35 struct ask{int k,n,rnk;}q[200010]; 36 bool operator <(ask a,ask b){return a.n<b.n;} 37 int cnt[1324]; 38 int ans[100010]; 39 inline void init() 40 { 41 memset(isprime,1,sizeof(isprime)); 42 for (int i=1;i<=100000;i++)sum[i]=1; 43 isprime[0]=isprime[1]=0; 44 int k,l; 45 for (int i=2;i<=100000;i++) 46 { 47 if (isprime[i]) 48 { 49 sum[i]=3; 50 for (int j=2;i*j<=100000;j++) 51 { 52 isprime[i*j]=0;k=1;l=j; 53 while (l%i==0){l/=i;k++;} 54 sum[i*j]*=2*k+1; 55 } 56 } 57 } 58 } 59 int main() 60 { 61 init(); 62 n=read(); 63 for (int i=1;i<=n;i++) 64 { 65 int x=read(),y=ceil(sqrt(read())),z=floor(sqrt(read())); 66 mx=max(mx,z); 67 q[2*i-1].k=x;q[2*i-1].n=y-1;q[2*i-1].rnk=-i; 68 q[2*i].k=x;q[2*i].n=z;q[2*i].rnk=i; 69 } 70 sort(q+1,q+2*n+1); 71 int now=1; 72 for (int i=0;i<=mx;i++) 73 { 74 cnt[sum[i]]++; 75 while (now<=2*n&&q[now].n==i) 76 { 77 if (q[now].rnk>0)ans[q[now].rnk]+=cnt[q[now].k]; 78 else ans[-q[now].rnk]-=cnt[q[now].k]; 79 now++; 80 } 81 } 82 for (int i=1;i<=n;i++)printf("%d ",ans[i]); 83 }