HDU-1028-Ignatius and the Princess III

HDU-1028-Ignatius and the Princess III

Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 1028

Integer Division, helpless, timeout

# Include <stdio. h> int huafen (int n, int m) // divide n into the maximum number of integers not greater than m {if (n <1 | m <1) return 0; if (n = 1 | m = 1) return 1; if (n <m) return huafen (n, n); if (n = m) return huafen (n m-1) + 1; return huafen (m-1) + huafen (n-m, m ); // The maximum addition = m and the maximum addition <= composition of the analysis result of the analysis.} int main () {int t; while (scanf ("% d", & t )! = EOF) printf ("% d \ n", huafen (t, t); return 0 ;}

Referring to TankyWoo's blog, I used the primary function to solve the problem of integer division.

Example

Each weight has one weight of one gram, two grams, three grams, and four grams. What kind of weight can be claimed? What are the possible solutions for each weight?

One 1 gram weight can be represented by the function 1 + x

1 2 grams of weight can be expressed by the function 1 + x ^ 2

1 3 grams of weight can be represented by the function 1 + x ^ 3

1 4 grams of weight can be represented by the function 1 + x ^ 4

(1 + x) (1 + x ^ 2) (1 + x ^ 3) (1 + x ^ 4)

= (1 + x ^ 2 + x ^ 3) (1 + x ^ 3 + x ^ 4 + x ^ 7)

= 1 + x ^ 2 + 2x ^ 3 + 2x ^ 4 + 2x ^ 5 + 2x ^ 6 + 2x ^ 7 + x ^ 8 + x ^ 9 + x ^ 10

For example, there are 2x ^ 5 items on the right end, that is, the 5 grams of the solution are 2: 5 = 3 + 2 = 4 + 1; 6 = 1 + 2 + 3 = 4 + 2; 10 = 1 + 2 + 3 + 4.

Therefore, there are 2 solutions for 6 grams and 1 solutions for 10 grams.

Thus, we can construct Integer Division functions.

G (x) = (1 + x ^ 2 + ......) * (1 + x ^ 2 + x ^ 4 .....) * (1 + x ^ 3 + x ^ 6 .......) *.......

Just multiply the polynomial to obtain the coefficient of x ^ n.

#include<stdio.h>int c1[10001],c2[10001];int main(){int n,i,j,k;while(scanf("%d",&n)!=EOF){for(i=0;i<=n;i++){c1[i]=1;c2[i]=0;}for(i=2;i<=n;i++){for(j=0;j<=n;j++)for(k=0;k+j<=n;k+=i)c2[j+k]+=c1[j];for(j=0;j<=n;j++){c1[j]=c2[j];c2[j]=0;}}printf("%d\n",c1[n]);}return 0;}