poj 1579 Function Run Fun

http://162.105.81.212/JudgeOnline/problem?id=1579

刚开始题目最后一句话没看,

真写了个递归函数w(a,b,c)囧,

结果50 50 50那组数据就过不了.

于是就改用dp

dp转换方程就不用说了,就是题目给的(那么多与呀或呀的,别看错了,- -||)

时间复杂度O(n^3)n<=20

#include<iostream>
using namespace std;
int main()
{
int a, b, c, dp[21][21][21], big;
while(scanf("%d%d%d", &a, &b, &c) != EOF)
{
big = 0;
if(a==-1 && b==-1 && c==-1)
break;
if(a<=0 || b<=0 || c<=0)
{
printf("w(%d, %d, %d) = %d/n", a, b, c, 1);
continue;
}
if(a>20 || b>20 || c>20)
{
big = 1;
printf("w(%d, %d, %d) = ", a, b, c);
a = 20, b = 20, c = 20;
}
for(int i=0; i<=a; i++)
for(int j=0; j<=b; j++)
for(int k=0; k<=c; k++)
{
if(i<=0 || j<=0 || k<=0)
dp[i][j][k] = 1;
else if(i<j && j<k)
dp[i][j][k] = dp[i][j][k-1] + dp[i][j-1][k-1] - dp[i][j-1][k];
else
dp[i][j][k] = dp[i-1][j][k] + dp[i-1][j-1][k] + dp[i-1][j][k-1] - dp[i-1][j-1][k-1];
}
if(big)
printf("%d/n", dp[a][b][c]);
else
printf("w(%d, %d, %d) = %d/n", a, b, c, dp[a][b][c]);
}
return 0;
}