Codeforces Round #240 (Div. 1) B. Mashmokh and ACM DP
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007(109 + 7).
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
3 2
5
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
题意:给你一个N,K, 表示从1到n,选取长度为k的序列A满足 Ai整除Ai-1
题解:dp[i][j]表示长度为j是最大为i的序列方案数
//1085422276 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<queue> #include<cmath> #include<map> #include<bitset> #include<set> #include<vector> using namespace std ; typedef long long ll; #define mem(a) memset(a,0,sizeof(a)) #define meminf(a) memset(a,127,sizeof(a)); #define memfy(a) memset(a,-1,sizeof(a)) #define TS printf("111111\n"); #define FOR(i,a,b) for( int i=a;i<=b;i++) #define FORJ(i,a,b) for(int i=a;i>=b;i--) #define READ(a,b,c) scanf("%d%d%d",&a,&b,&c) #define mod 1000000007 #define maxn 2005 inline ll read() { ll x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') { if(ch=='-')f=-1; ch=getchar(); } while(ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); } return x*f; } //**************************************** ll dp[maxn][maxn]; int main() { int n=read(),m=read(); mem(dp); FOR(i,1,n)dp[i][1]=1; FOR(k,1,m-1) FOR(i,1,n) { if(dp[i][k]) for(int j=i;j<=n;j+=i) { dp[j][k+1]=(dp[j][k+1]+dp[i][k])%mod; } } ll ans=0; FOR(i,1,n)ans=(ans+dp[i][m])%mod; cout<<ans<<endl; return 0; }