Sicily 1029. Rabbit 解题报告

题目传送门：1029. Rabbit

思路：

　　题目说的有一点奇怪，兔子要过m个月才能长大，但是第m个月的时候已经可以生孩子了，这是需要注意的。

　　思路也比较简单，用一个rabbits[months_to_grow + 1]的数组存储各个月年龄的兔子数量，其中刚出生的存储在下标0中，成年的存储在下标months_to_grow中。显然下个月的时候rabbit[0] = rabbit[months_to_grow],并且其他年龄段的兔子年龄会向前推进，直到成年之后就一直在rabbit[months_to_grow]中。

　　看到样例 1267650600228229401496703205376 的输出，只能用string类型来存储兔子数量，进行加法的时候随便写了一个模拟手算的加法，无奈不懂什么高精度加法。最后把各年龄段数量加起来就是结果。

代码：

#include<iostream>
using namespace std;

string BigIntAdd(string &s1,string &s2);

int main(){
    int months_to_grow,months;
    while(cin >> months_to_grow >> months
            && (months_to_grow != 0 || months != 0)){
        string rabbits[months_to_grow + 1],result = "0";//use string to store BigInt
        rabbits[months_to_grow] = "1";//the number of adult rabbits stored in rabbits[months_to_grow]
        for(int i = 0;i < months_to_grow;i++)
            rabbits[i] = "0";
        for(int i = 0;i < months;i++){
            rabbits[months_to_grow] = BigIntAdd(rabbits[months_to_grow] , rabbits[months_to_grow - 1]);
            for(int j = months_to_grow - 1;j > 0;j--){
                rabbits[j] = rabbits[j - 1];
            }
            rabbits[0] = rabbits[months_to_grow];
        }
        for(int i = 0;i <= months_to_grow;i++)//add the rabbits of all months to get total rabbits
            result = BigIntAdd(result,rabbits[i]);
        cout << result << endl;
    }
    return 0;
}
string BigIntAdd(string &s1,string &s2){
    //Pre: given two integers represented as string
    //Post:the sum of the two integers will be calculated and returned,the process is like
    //        calculating by hand.
    int len1 = s1.length(),len2 = s2.length();
    int length,carry = 0;//length is the length of result.
    if(len1 < len2){
        length = len2;
        for(int i = 0;i < len2 - len1;i++) //Add prefix 0 to s1
            s1 = "0" + s1;
    }else{
        length = len1;
        for(int i = 0;i < len1 - len2;i++) //Add prefix 0 to s2
            s2 = "0" + s2;
    }
    string result(length,'0');
    for(int i = length - 1;i >= 0;i--){
        if(s1[i] - '0' + s2[i] - '0' + carry < 10){
            result[i] = s1[i] - '0' + s2[i] + carry;
            carry = 0;
        }else{
            result[i] = s1[i] - '0' + s2[i] - 10 + carry;
            carry = 1;
        }
    }
    if(carry == 1)
        result = "1" + result;
    return result;
}