zoukankan      html  css  js  c++  java
  • (动态规划)cf 431C

    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.

    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: nk 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

    神奇的动态规划啊啊啊啊啊啊,

    两个状态,dp[n][0,1] 0,1代表是否选超过d的。

    dp[i+j][0]+=dp[i][0] j<d

    dp[i+j][1]+=dp[i][0] j>=d

    dp[i+j][1]+=dp[i][1]

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<string>
    #include<algorithm>
    using namespace std;
    #define LL long long
    #define MOD 1000000007
    LL n,k,d,dp[110][2];
    int main()
    {
          scanf("%I64d%I64d%I64d",&n,&k,&d);
          memset(dp,0,sizeof(dp));
          dp[0][0]=1;
          for(int i=0;i<n;i++)
          {
                for(int j=1;j<=k;j++)
                {
                      if(i+j>n) continue;
                      if(j<d) dp[i+j][0]+=dp[i][0];
                      else dp[i+j][1]+=dp[i][0];
                      dp[i+j][1]+=dp[i][1];
                      dp[i+j][0]=dp[i+j][0]%MOD;
                      dp[i+j][1]=dp[i+j][1]%MOD;
                }
          }
          printf("%I64d",dp[n][1]);
          return 0;
    }
    

      

  • 相关阅读:
    堆排序
    剑指 Offer 59
    面试题:happen-before原则和as-if-serial语义
    面试题:Redis的持久化机制是什么?各自的优缺点?
    面试题:单线程redis还这么快
    面试题:微服务理论
    wait和notify
    线程八锁
    面试题:在静态方法和非静态方法上加 Synchronized的区别
    面试题:3种线程阻塞唤醒的对比
  • 原文地址:https://www.cnblogs.com/a972290869/p/4242543.html
Copyright © 2011-2022 走看看