CF 431C k-Tree

C. k-Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a k-tree.

A k-tree is an infinite rooted tree where:

• each vertex has exactly k children;
• each edge has some weight;
• if we look at the edges that goes from some vertex to its children (exactly k edges), then their weights will equal 1, 2, 3, ..., k.

The picture below shows a part of a 3-tree.

As soon as Dima, a good friend of Lesha, found out about the tree, he immediately wondered: "How many paths of total weight n (the sum of all weights of the edges in the path) are there, starting from the root of a k-tree and also containing at least one edge of weight at least d?".

Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo1000000007 (109 + 7).

Input

A single line contains three space-separated integers: n, k and d (1 ≤ n, k ≤ 100; 1 ≤ d ≤ k).

Output

Print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

Sample test(s)

input

3 3 2

output

3

input

3 3 3

output

1

input

4 3 2

output

6

input

4 5 2

output

7

题意：
k-Tree，每个节点有k个儿子，边权分别是1、2、……、K。
现求出从根节点出发，有多少条路径，使得总权值恰好为N，并且每条路径上至少有一条权值不少于d的边。
分析：
      dp[i][x]表示当到达i状态时的总数，x为0或1，0表示之前没有使用过>=d的边，1表示之前有使用过>d的边。
则有状态转移方程：
                                               dp[i][0] += dp[i-j][0] , (j < d)
                                               dp[i][1] += dp[i-j][0] , (j>=d)
                                               dp[i][1] += dp[i-j][1]
初始化dp[0][0] = 1;
这题还需要注意数据的类型，不能写成INT，应该写成LONG LONG。。。。。。。。。。这会导致WA。（注意）
#include<iostream>#include<cstdio>#include<cstring>long long dp[200][2];using namespace std;long long mod = 1000000007;int main(){long long n,k,d;while(~scanf("%I64d%I64d%I64d",&n,&k,&d)){memset(dp,0,sizeof(dp));dp[0][0] = 1;for(int i=1;i<=n;i++){for(int j=1;j<=k;j++){if(j<d&&i-j>=0){dp[i][0] += dp[i-j][0];dp[i][1] += dp[i-j][1];}else if(j>=d&&i-j>=0){dp[i][1] += dp[i-j][0];dp[i][1] += dp[i-j][1];}dp[i][0] = dp[i][0]%mod;dp[i][1] = dp[i][1]%mod;}}cout<<dp[n][1]<<endl;}return 0;}