codeforces 261E Maxim and Calculator

Maxim has got a calculator. The calculator has two integer cells. Initially, the first cell contains number 1, and the second cell contains number 0. In one move you can perform one of the following operations:

Let's assume that at the current time the first cell contains number a, and the second cell contains number b. Write to the second cell number b + 1; Let's assume that at the current time the first cell contains number a, and the second cell contains number b. Write to the first cell number a·b.

Maxim is wondering, how many integers x (l ≤ x ≤ r) are there, such that we can write the number x to the first cell of the calculator, having performed at most p moves.

Input

The first line contains three integers: l, r, p (2 ≤ l ≤ r ≤ 109, 1 ≤ p ≤ 100).

The numbers in the line are separated by single spaces.

Output

In a single line print a single integer — the answer to the problem.

Examples input 2 10 3 output 1 input 2 111 100 output 106 input 2 111 11 output 47 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

DP+思路~

神奇的DP方法~以步数为依据DP

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,l,r,p,tot,a[4000001],pri[100],f[4000001],ans; bool bb[100],b[4000001]; void dfs(int u,int v) { a[++n]=u; for(int i=v;i<=tot;i++) if(u>r/pri[i]) return; else dfs(u*pri[i],i); } int main() { scanf("%d%d%d",&l,&r,&p); for(int i=2;i<p;i++) { if(!bb[i]) pri[++tot]=i; for(int j=1;j<=tot && pri[j]*i<p;j++) { bb[i*pri[j]]=1; if(!(i%pri[j])) break; } } dfs(1,1); sort(a+1,a+n+1); memset(f,127/2,sizeof(f)); f[1]=0; for(int i=2;i<p;i++) for(int j=1,k=1;j<=n;j++) if(!(a[j]%i)) { while(a[k]*i<a[j]) k++; f[j]=min(f[j],f[k]+1); if(f[j]+i<=p) b[j]=1; } for(int i=1;i<=n;i++) if(b[i] && a[i]>=l) ans++; printf("%d\n",ans); return 0; }