兔子
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Problem 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.
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 33 51 1000 0
Sample Output
591267650600228229401496703205376
使用大数模板
int main(){ //freopen("b.txt","r",stdin); int d,m,i;BigNum r[105];while(scanf("%d %d",&m,&d)==2&&d+m) { for(i=1;i<=m;i++) r[i]=i+1; //当m<=d时,成年兔子数目不变,每月产一对for(i=m+1;i<=d;i++)r[i]=r[i-1]+r[i-m]; //当m>d之后,当月兔子总数等于上月兔子总数加上m个月前的兔子总数r[d].print(); }return 0;}
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