递归的函数
Problem Description
Input
Output
Sample Input
1 1 1 2 2 2
Sample Output
2 4
//此题若利用递归则会超时
// 所以采用动态规划
#include<stdio.h>
int t[100][100][100] = {{{0}}}; //必须初始化数组
int f(int a, int b, int c)
{
if(a <= 0 || b <= 0 || c <= 0)
{
return 1;
}
else if(a > 20 || b > 20 || c > 20)
{
return f(20, 20, 20); //这个也是一个返回条件
}
else if(a < b && b < c)
{
if(t[a][b][c] == 0) //如果为零则说明还未进行处理
{
return t[a][b][c] = f(a, b, c - 1) + f(a, b - 1, c - 1) - f(a, b - 1, c);
}
else
{
return t[a][b][c];
}
}
else
{
if(t[a][b][c] == 0)
{
return t[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);
}
else
{
return t[a][b][c];
}
}
}
int main(void)
{
int a, b, c, m;
while(~scanf("%d %d %d", &a, &b, &c))
{
m = f(a, b, c);
printf("%d\n", m);
}
return 0;
}