zoukankan      html  css  js  c++  java
  • UVa11137

     Ingenuous Cubrency

    People in Cubeland use cubic coins. Not only the unit of currency is called acube but also the coins are shaped like cubes and their values are cubes. Coins with values of all cubic numbers up to 9261 (= 213), i.e., coins with the denominations of 1, 8, 27, ..., up to 9261 cubes, are available in Cubeland.

    Your task is to count the number of ways to pay a given amount using cubic coins of Cubeland. For example, there are 3 ways to pay 21 cubes: twenty one 1cube coins, or one 8 cube coin and thirteen 1 cube coins, or two 8 cube coin and five 1 cube coins.

    Input consists of lines each containing an integer amount to be paid. You may assume that all the amounts are positive and less than 10000.

    For each of the given amounts to be paid output one line containing a single integer representing the number of ways to pay the given amount using the coins available in Cubeland.

    Sample input

    10 
    21
    77
    9999

    Output for sample input

    2
    3
    22
    440022018293

    题意:

           输入正整数n(n <= 10000),求将n写成若干个正整数的立方和有多少种方式。

    分析:

    建立多段图。结点(i,j)表示“使用不超过i的正整数的立方,累加和为j”这个状态,设dp(i,j)为从(0,0)到(i,j)的路径的条数,最终答案为dp(21,n)(22*22*22>n)。

    转移方程是dp[i][j + a*i*i*i] += dp[i-1][j] (i + a*i*i*i<=n)

     1 #include <cstdio>
     2 #include <cstring>
     3 const int maxn = 10000;
     4 int n;
     5 long long dp[22][maxn + 1];
     6 int main(){
     7     dp[0][0] = 1;
     8     for(int i = 1 ; i <= 21 ; i++)for(int j = 0 ; j <= maxn ; j++)
     9         for(int a = 0 ; a * i * i * i + j <= maxn ; a++)
    10             dp[i][j + a * i * i * i] += dp[i - 1][j];
    11     while(~scanf("%d",&n)) printf("%lld
    ", dp[21][n]);
    12     return 0;
    13 }
    View Code
  • 相关阅读:
    带外数据
    数组中的第K个最大元素
    广播和多播
    反转链表
    ioctl操作
    非阻塞式I/O
    [CSP-S模拟测试]:简单的括号序列(组合数)
    [CSP-S模拟测试]:最大异或和(数学)
    关于我
    [CSP-S模拟测试]:礼物(数学)
  • 原文地址:https://www.cnblogs.com/cyb123456/p/5804251.html
Copyright © 2011-2022 走看看