Poj 3982 序列

1.Link：

http://poj.org/problem?id=3982

2.Content：

序列

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7057		Accepted: 3182

Description

数列A满足An = An-1 + An-2 + An-3, n >= 3

编写程序，给定A0, A1 和 A2, 计算A99

Input

输入包含多行数据

每行数据包含3个整数A0, A1, A2 (0 <= A0, A1, A2 <= 32767) 
数据以EOF结束

Output

对于输入的每一行输出A99的值

Sample Input

1 1 1

Sample Output

69087442470169316923566147

Source

3.Method：

套用大数相加模板

http://www.cnblogs.com/mobileliker/p/3512632.html

4.Code：

 #include <iostream>
 #include <string>

 using namespace std;

 ;

 string sum(string s1,string s2)
 {
     if(s1.length()<s2.length())
     {
         string temp=s1;
         s1=s2;
         s2=temp;
     }
     int i,j;
     ,j=s2.length()-;i>=;i--,j--)
     {
         s1[i]=?s2[j]-));   //注意细节
         )
         {
             s1[i]=+');
             ]++;
             '+s1;
         }
     }
     return s1;
 }

 int main()
 {
     //freopen("D://input.txt","r",stdin);

     int i;

     string str0,str1,str2,str3;

     while(cin >> str0 >> str1 >> str2)
     {
         //cout << str0 << endl;
         //cout << str1 << endl;
         //cout << str2 << endl;

         i = Num - ;
         while(i--)
         {
             str3 = sum(str0,str1);
             str3 = sum(str3,str2);
             str0 = str1;
             str1 = str2;
             str2 = str3;
         }

         cout << str3 << endl;

     }

     ;
 }

5.Reference: