Sicily 1029. Rabbit

1029. Rabbit
      Constraints
      Time Limit: 1 secs, Memory Limit: 32 MB
      Description
      The rabbits have powerful reproduction ability. One pair of adult rabbits can give birth to one pair of kid rabbits every month. And after m months, the kid rabbits can become adult rabbits.
      As we all know, when m=2, the sequence of the number of pairs of rabbits in each month is called Fibonacci sequence. But when m<>2, the problem seems not so simple. You job is to calculate after d months, how many pairs of the rabbits are there if there is exactly one pair of adult rabbits initially. You may assume that none of the rabbits dies in this period.
      Input
      The input may have multiple test cases. In each test case, there is one line having two integers m(1<=m<=10), d(1<=d<=100), m is the number of months after which kid rabbits can become adult rabbits, and d is the number of months after which you should calculate the number of pairs of rabbits. The input will be terminated by m=d=0.
      Output
      You must print the number of pairs of rabbits after d months, one integer per line.
      Sample Input
      2 3
      3 5
      1 100
      0 0
      Sample Output
      5
      9
      1267650600228229401496703205376
      這道題是一個動態規劃問題，可以找到遞推關係式，設a[n]爲第n個月的兔子總數,則a[n]=a[n-1]+a[n-m]，a[n-1]是上個月的兔子數量，a[n-m]是能生產的兔子數量，即這個月新出生的兔子數量，然後這道題還涉及大整數加法問題，其核心思想就是用數組存儲大整數，從低位向高位依次相加，有進位的話就加到前一位，然後對本位數字進行對10的取模，核心代碼如下
      

#include<iostream>
#include<cstring>
using namespace std;
int main(){
	int m,d;
	int a[105][105];
	while(cin>>m>>d&&m!=0&&d!=0){
		memset(a,0,sizeof(a));
		a[1][104]=2;
		for(int i=2;i<=m;i++){
			a[i][104]=i+1;
		}
		int now=0;
		for(int i=m+1;i<=d;i++){
			for(int j=104;j>=0;j--){
				a[i][j]=a[i-1][j]+a[i-m][j]+a[i][j];
				if(a[i][j]>9){
					a[i][j-1]+=a[i][j]/10;
					a[i][j]=a[i][j]%10;
				}
			}
		}
		for(int i=0;i<105;i++){
			if(a[d][i]!=0){
				for(int j=i;j<105;j++){
					cout<<a[d][j];	
				}
				break;
			}
		}
		cout<<endl;
	}
}