题目链接:https://cn.vjudge.net/contest/269853#problem
题目描述:
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.
Input
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.
Output
Print the value for w(a,b,c) for each triple.
Sample Input
1 1 1 2 2 2 10 4 6 50 50 50 -1 7 18 -1 -1 -1
Sample Output
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
思路:
这里主要是用的递归,但若是每次计算都要递归一次的话,最后会超时,所以用一个三维数组将结果先存起来,使用时直接调用结果即可,不需要一次次的重复递归运算。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1010;
int dp[25][25][25];
int fun(int a,int b,int c)
{
if(a<=0||b<=0||c<=0)
return 1;
else if(a>20||b>20||c>20)
return dp[20][20][20];
else
return dp[a][b][c];
}
void yu()//用预处理
{
for(int i=1;i<=20;i++)
for(int j=1;j<=20;j++)
for(int k=1;k<=20;k++)
{
if(i<j&&j<k)
dp[i][j][k]=fun(i,j,k-1)+fun(i,j-1,k-1)-fun(i,j-1,k);
else
dp[i][j][k]=fun(i-1,j,k)+fun(i-1,j-1,k)+fun(i-1,j,k-1)-fun(i-1,j-1,k-1);
}
}
int main()
{
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
yu();
if(a==-1&b==-1&&c==-1)break;
if(a>20&&b>20&&c>20)
printf("w(%d, %d, %d) = 1048576\n",a,b,c);
else
{
int sum=fun(a,b,c);
printf("w(%d, %d, %d) = %d\n",a,b,c,sum);
}
}
return 0;
}