codeforces-431C-k-Tree【dp】

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codeforces-431C-k-Tree【dp】

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                time limit per test1 second     memory limit per test256 megabytes

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 modulo 1000000007 (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).

input
3 3 2
output
3

input
3 3 3
output
1

input
4 3 2
output
6

input
4 5 2
output
7

题目链接：cf-431C

题目大意：满k叉树，每个顶点有k条边，边的权重为1~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。

题目思路：dp[i][h],第一个参数从1-n，遍历权值为i的情况，h有两种情况，0和1,0表示权值为i的所有情况，1表示权值为i的时候路径中存在一个值大于d的情况。

状态转移方程：

//权值为i的处理所有情况

    dp[i][0] = dp[i][0] + dp[i - j][0]

//处理权值为i且存在边大于等于d的情况

        /* if (j >= d) dp[i][1] = dp[i][1] + dp[i - j ][0]   

           else dp[i][1] = dp[i][1] + dp[i - j ][1]        */

        即：dp[i][1] = dp[i][1] + dp[i - j][j

所以结果为dp[n][1]

以下是代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int dp[105][105];
#define mod 1000000007
int main(){
    int n,k,d;
    cin >> n >> k >> d;
    dp[0][0] = 1;               
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= min(k,i); j++)
        {
            dp[i][0] = (dp[i][0] + dp[i - j][0]) % mod;
            dp[i][1] = (dp[i][1] + dp[i - j][jcout << (dp[n][1] % mod) << endl;
    return 0;
}