2013
12-09

# Function Run Fun

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.

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.

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

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

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;
}

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;