首页 > ACM题库 > HDU-杭电 > HDU 1331 Function Run Fun-递归和分治-[解题报告] C++
2013
12-09

HDU 1331 Function Run Fun-递归和分治-[解题报告] C++

Function Run Fun

问题描述 :

We all love recursion! Don’t we?

Consider a three-parameter recursive function w(a, b, c):

if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns:
1

if a > 20 or b > 20 or c > 20, then w(a, b, c) returns:
w(20, 20, 20)

if a < b and b < c, then w(a, b, c) returns:
w(a, b, c-1) + w(a, b-1, c-1) – w(a, b-1, c)

otherwise it returns:
w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) – w(a-1, b-1, c-1)

This is an easy function to implement. The problem is, if implemented directly, for moderate values of a, b and c (for example, a = 15, b = 15, c = 15), the program takes hours to run because of the massive recursion.

输入:

The input for your program will be a series of integer triples, one per line, until the end-of-file flag of -1 -1 -1. Using the above technique, you are to calculate w(a, b, c) efficiently and print the result.

输出:

Print the value for w(a,b,c) for each triple.

样例输入:

1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1 

样例输出:

w(1, 1, 1) = 2
w(2, 2, 2) = 4
w(10, 4, 6) = 523
w(50, 50, 50) = 1048576
w(-1, 7, 18) = 1

#include"iostream"
using namespace std;

int f[25][25][25];

void w(int a,int b,int c)
{
    if(a<=0||b<=0||c<=0)
    {
        f[a][b][c]=1;
    }
    else if(a<b&&b<c)
    {
        f[a][b][c]=f[a][b][c-1]+f[a][b-1][c-1]-f[a][b-1][c];
    }
    else
    {
        f[a][b][c]=f[a-1][b][c]+f[a-1][b-1][c]+f[a-1][b][c-1]-f[a-1][b-1][c-1];
    }
}

int main()
{
    int a,b,c;
    int i,j,k;
    for(i=0;i<21;i++)
        for(j=0;j<21;j++)
            for(k=0;k<21;k++)
                w(i,j,k);
    while(cin>>a>>b>>c)
    {
        if(a==-1&&b==-1&&c==-1)
            break;
        if(a<=0||b<=0||c<=0)
        {
            cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<1<<endl;
            continue;
        }
        else if(a>20||b>20||c>20)
        {
            cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<f[20][20][20]<<endl;
            continue;
        }
        cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<f[a][b][c]<<endl;
    }
    return 0;
}

解题报告转自:http://www.cnblogs.com/Action-/archive/2012/04/18/2454654.html


  1. 第23行:
    hash = -1是否应该改成hash[s ] = -1

    因为是要把从字符串s的start位到当前位在hash中重置

    修改提交后能accept,但是不修改居然也能accept

  2. int half(int *array,int len,int key)
    {
    int l=0,r=len;
    while(l<r)
    {
    int m=(l+r)>>1;
    if(key>array )l=m+1;
    else if(key<array )r=m;
    else return m;
    }
    return -1;
    }
    这种就能避免一些Bug
    l,m,r
    左边是l,m;右边就是m+1,r;