2nd JOJ Cup Online VContest Problem
We all know that bunny is fond of carrots. One cloudy day, he was told that there would be a violenting flood coming soon to destroy the forests. He was scared and after not thinking too much he told himself that he had to escape. He suddenly recalled that there was a temple on the top of the hill and he might shelter there until the flood's past. But unfortunately there was no food for him on the top of the hill, so he had to take his carrots away along with himself. Then he moved to the foot of the hill and stopped. There was only one way for him to get the top of the hill, that is, a long staircase. Given the number of the steps of the staircase, he asked himself:"how many different ways of strides are there for him to get the top of the hill?". Of course, because of his height, he could only stride a limited range of steps. He was smart so much so that he got the answer quickly. Do you know how he did it?
Input Specification
The input consists of several test cases, each of which occupies a line containing M(1<=M<=40) and N(1<=N<=10), where M indicates the number of the steps of the staircase and N indicates the maximal number of steps the bunny can stride once.
Output Specification
Each test case should correspond to a line in the output. Your program should print an integer which is the answer.
Sample Input
4 2
5 4
Sample Output
5
15
題意是一只兔子要到距離為M（單位為1）的地方，它每步最多走N，問有多少種方法。輸入M N 輸出r[m][n];
解析：設(shè)為r[i][j]，表示距離總共i且每次最多走j的方法數(shù)(可以沒有走出那大小為j的那步，只是允許走那步而已)。
根據(jù)最后那一步可能走的長度，r[i-2][j]表示最后那步距離為2.r[i-j][j]表示最后那步距離為j.
建立遞推關(guān)系r[i][j]=r[i-1][j]+r[i-2][j]+r[i-3][j]+r[i-j][j];
//本人覺得這里特別難想，想到了也覺得無法建立遞推關(guān)系呀，j不是一直不變嗎？
神奇的地方在于可以令r[0][j]=r[1][j]=1;當(dāng)然還有r[i][1]=1
-----------預(yù)處理----
for(j=0;j<n;j++)
{
r[0][j]=1;
r[1][j]=1;
}
for(i=0;i<n;i++)
r[i][1]=1;
---dp----------
for(i=2;i<m;i++)
for(j=2;j<n;j++)
{
if(i<j)
r[i][j]=r[i][i]; //這點(diǎn)非常重要
else
for(k=1;k<=j;k++)
{
r[i][j]+=r[i-k][j];
}
}
------------
r[1][1]=1
r[2][1]=1
r[2][2]=r[2][1]+r[2][2]=2;
r[3][1]=1;
r[3][2]=r[2][2]+r[1][2]=3;
r[3][3]=r[0][3]+r[1][3]+r[2][3]=4;
posted on 2009-07-20 17:19
luis 閱讀(564)
評論(0) 編輯 收藏 引用 所屬分類:
動態(tài)規(guī)劃