Sicily 1029. Rabbit

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. inputthe 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 shoshould calculate the number of pairs of rabbits. the input will be terminated by M = D = 0. outputyou must print the number of pairs of Rabbits after D months, one integer per line. sample input copy sample input to clipboard

2 33 51 1000 0

Sample output

591267650600228229401496703205376

 

Analysis: first, the recursive formula F (n) = f (n-1) + f (n-m) can be obtained through analysis. Then, the above recursive formula is used to obtain F (d). In general, recursion is not required, because recursion consumes a lot of time when the value is large, so here we use the memory method, in fact, to save the results with an array. Because the topic describes d <= 100, the number of results can be saved in an array of 100. The other one is the big integer calculation. Here I use a string to represent it. Although it looks a bit annoying, it can still be understood.

 

# Include <iostream> # include <string> # include <algorithm> using namespace STD; string add (string num1, string num2) {// high-precision addition reverse (num1.begin (), num1.end (); reverse (num2.begin (), num2.end (); string: iterator iter1 = num1.begin (); string: iterator iter2 = num2.begin (); string STR = ""; int add = 0; For (; iter1! = Num1.end (), iter2! = Num2.end (); ++ iter1, ++ iter2) {int value = (* iter1-'0') + (* iter2-'0') + Add; STR = (char) (Value % 10) + '0') + STR; add = value/10;} string: iterator iter = (iter1 = num1.end ()? Iter2: iter1); string: iterator iter_end = (iter1 = num1.end ()? Num2.end (): num1.end (); For (; iter! = Iter_end; ++ ITER) {int value = (* ITER-'0') + Add; STR = (char) (Value % 10) + '0 ') + STR; add = value/10;} If (add! = 0) STR = (char) (add + '0') + STR; return STR;} string getresult (int m, int D) {// The total number of recursive rabbits string result [103]; for (INT I = 0; I <= D; I ++) {if (I <= m) {int value = I + 1; char T [256]; sprintf (T, "% d", value); Result [I] = string (t );} else result [I] = add (result [I-1], result [I-m]); // recursive formula: F (n) = f (n-1) + f (n-m)} return result [d];} int main (INT argc, char const * argv []) {int M, D; whi Le (CIN> m> D & M! = 0 & D! = 0) {cout <getresult (M, d) <Endl;} return 0 ;}

 

 

Sicily 1029. Rabbit