hrbust Function Run Fun

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这题就怕te

递归可以

也可以直接迭代出所有情况

关系式别看错了就好

Function Run FunTime Limit: 1000 MSMemory Limit: 65536 KTotal Submit: 380(184 users)Total Accepted: 198(170 users)Rating: Special Judge: NoDescription

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 Input1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1Sample Outputw(1, 1, 1) = 2
w(2, 2, 2) = 4
w(10, 4, 6) = 523
w(50, 50, 50) = 1048576
w(-1, 7, 18) = 1

解法一（递归并记录路径）

#include<stdio.h>#include<string.h>int re[25][25][25];int w(int a,int b,int c){   if (a <= 0 || b <= 0 || c <= 0)    return 1;    if(re[a][b][c])        return re[a][b][c];    if(a > 20 | b > 20 | c > 20)    return w(20,20,20);   if(a < b && b < c)    return re[a][b][c]=w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c) ;    return re[a][b][c]=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(){    int n,m,z;    memset(re,0,sizeof(re));    while(~scanf("%d%d%d",&n,&m,&z))    {        if(n==-1&&m==-1&&z==-1)break;        if(n<=0||m<=0||z<=0)            printf("w(%d, %d, %d) = 1\n",n,m,z);        else if(n>20||m>20||z>20)            printf("w(%d, %d, %d) = %d\n",n,m,z,w(20,20,20));        else printf("w(%d, %d, %d) = %d\n",n,m,z,w(n,m,z));    }}

解法二迭代

#include<stdio.h>int a[25][25][25];int main(){    for(int i=0; i<=20; i++)    {        for(int j=0; j<=20; j++)        {            for(int k=0; k<=20; k++)            {                if(i==0||j==0||k==0)                {                    a[i][j][k]=1;                    continue;                }                if(i<j&&j<k)                    a[i][j][k]=a[i][j][k-1]+a[i][j-1][k-1]-a[i][j-1][k];                else a[i][j][k]= a[i-1][j][k]+a[i-1][j-1][k]+a[i-1][j][k-1]-a[i-1][j-1][k-1];                // printf("%d ",a[i][j][k]);            }          //  printf("\n");        }        //printf("\n");    }    int n,m,z;    while(~scanf("%d%d%d",&n,&m,&z))    {        if(n==-1&&m==-1&&z==-1)break;        if(n<=0||m<=0||z<=0)            printf("w(%d, %d, %d) = 1\n",n,m,z);        else if(n>20||m>20||z>20)            printf("w(%d, %d, %d) = %d\n",n,m,z,a[20][20][20]);        else printf("w(%d, %d, %d) = %d\n",n,m,z,a[n][m][z]);    }}

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