[sicily online]1029. Rabbit

/*1,大整數加法，用string類比2,用一個數組類比兔子成長過程ConstraintsTime Limit: 1 secs, Memory Limit: 32 MBDescriptionThe 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 should 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 Input2 33 51 1000 0Sample Output591267650600228229401496703205376*/#include<iostream>#include<fstream>#include<vector>#include<string>#include<algorithm>#include<cmath>#include<list>#include<map>#include<string.h>using namespace std;string strPlus(string x1,string x2){for(string::size_type i=0;i<x2.size();i++){if(i>=x1.size())x1.push_back('0');if(i>=x2.size())break;int num1=x1[i]-'0';int num2=x2[i]-'0';int sum=num1+num2;x1[i]=sum%10+'0';if(sum<10)continue;int j=i+1;while(1){if(j>=x1.size()){x1.push_back('1');break;}if(x1[j]=='9'){x1[j]='0';j++;}else{x1[j]++;break;}}}return x1;}int main(){int m,d;while(cin>>m>>d&&m!=0){vector<string> mouths(m);mouths[m-1]="1";while(d--){string tmpLast=mouths[m-1];if(m>1)mouths[m-1]=strPlus(mouths[m-1],mouths[m-2]);else mouths[m-1]=strPlus(mouths[m-1],mouths[m-1]);for(int i=m-3;i>=0;i--){mouths[i+1]=mouths[i];}//end forif(m>1)mouths[0]=tmpLast;}string sum="0";for(int i=0;i<m;i++)sum=strPlus(sum,mouths[i]);reverse(sum.begin(),sum.end());cout<<sum<<endl;}//end while}