【题解】ZOJ-1168 Function Run Fun

Function Run Fun (ZOJ - 1168)

题面

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
2
3
4
5
6
1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1

样例输出

1
2
3
4
5
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

提示

思路

代码

查看代码
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using namespace std;
typedef long long ll;

int ans[21][21][21];
int a, b, c;

int w(int a, int b, int c) {
if (a<=0 || b<=0 || c<=0)
return 1;

if (a>20 || b>20 || c>20)
return ans[20][20][20];

if (ans[a][b][c]>0)
return ans[a][b][c];

if (a<b && b<c)
return w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);

return w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1);
}

int main(void) {
for (int i=0; i<21; i++)
for (int j=0; j<21; j++)
for (int k=0; k<21; k++)
ans[i][j][k] = w(i, j, k);

while (scanf("%d%d%d", &a, &b, &c)==3) {
if (a==-1 && b==-1 && c==-1)
break;
printf("w(%d, %d, %d) = %d\n", a, b, c, w(a, b, c));
}
return 0;
}
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