hrbust Function Run Fun
2017-01-21 17:13
246 查看
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
解法一(递归并记录路径)
解法二 迭代
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 Fun | ||||||
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Description | ||||||
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 |
#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)); } } |
#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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