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.
跟之前作业的那题有点不一样，用long long也不够，必须写成大精度加法
一开始WA，发现大精度加法写错了，应该先加进位再取模而不是先取模再加，改掉之后AC，用时0.03s
View Code
1 #include<stdio.h> 2 #define MAX 101 3 int rabbit[MAX][MAX] = {1}; 4 void generate( int m, int d ); 5 void add( int des, int a, int b ); 6 7 int main() 8 { 9 int m, d; 10 11 while( scanf("%d %d", &m, &d) && m != 0 ) 12 { 13 generate(m, d); 14 } 15 return 0; 16 } 17 18 void add( int des, int a, int b ) 19 { 20 int i, temp, carry = 0; 21 22 for ( i = 0; i < MAX; i++ ) 23 { 24 temp = rabbit[a][i] + rabbit[b][i] + carry; 25 rabbit[des][i] = temp % 10; 26 carry = temp / 10; 27 } 28 29 } 30 31 void generate( int m, int d ) 32 { 33 int i, j; 34 35 for ( i = 1; i <= d; i++ ) 36 for ( j = 0; j < MAX; j++ ) 37 rabbit[i][j] = 0; 38 39 for ( i = 1; i <= d; i++ ) 40 { 41 if ( i < m ) 42 add( i, i - 1, 0 ); 43 else 44 add( i, i - 1, i - m ); 45 } 46 47 i = MAX - 1; 48 while( rabbit[d][i] == 0) 49 i--; 50 51 for( j = i; j >= 0; j-- ) 52 printf("%d", rabbit[d][j]); 53 54 printf("\n"); 55 }