Hat's Fibonacci（大数加法+直接暴力）

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1250

hdu1250：

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9442    Accepted Submission(s):
3096

Problem Description

A Fibonacci sequence is calculated by adding the
previous two members the sequence, with the first two members being both
1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) +
F(n-3) + F(n-4)
Your task is to take a number as input, and print that
Fibonacci number.

 

Input

Each line will contain an integers. Process to end of
file.

 

Output

For each case, output the result in a line.

 

Sample Input

100

 

Sample Output

4203968145672990846840663646

Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

 

Author

戴帽子的

 

题解，同一般的斐波那契数列来说，最先的我想到了用大数加法和快速幂矩阵的方法来做，但是由于大数乘法的模板是O(n^2)所以超时了，所以考虑只用大数加法的直接暴力方法，由于斐波那契数列增长很快，所以位数不超过2005的前提下最多n 不超过10000,然后就过了，希望看到的大神吗也可以帮我提供一个复杂度小的大数乘法的模板，万分感激

下面给出ac代码，和用矩阵快速幂tle的代码：

ac代码：

 #include <iostream>
 #include<cstdio>
 #include<string>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 string add(string s1,string s2)
 {
       string ans = "";
       int i,j,x,y,k=;
       for(i=s1.length()-,j=s2.length()-;i>= && j>= ;i--,j--)
       {
          x = s1[i] - '';
          y = s2[j] - '';
          ans += char((x+y+k)% + '');
          k = (x+y+k)/;
       }
       while(i>=)
       {
          x=s1[i]-'';
          ans += char ((x+k)% + '');
          k = (x+k)/;
          i--;
       }
       while(j>=)
       {
          y=s2[j]-'';
          ans += char((y+k)% + '');
          k = (y+k)/;
          j--;
       }
       if(k>)
       ans += '';
       //ans.reverse();
       reverse(ans.begin(),ans.end());
       return ans;
 }
 string ms1, ms2, ms3, ms4, ms;
 int main()
 {
     int n;
     while(~scanf("%d",&n))
     {
         ms1 = ms2 = ms3 = ms4 = "";
         if(n <= ) {puts(""); continue;}
         for(int i = ; i < n-; i++)
         {
             ms = add(ms1, ms2);
             ms = add(ms, ms3);
             ms = add(ms, ms4);
             ms1 = ms2;
             ms2 = ms3;
             ms3 = ms4;
              ms4 = ms;
         }
         cout << ms << endl;
   }
      return ;
 }

矩阵快速幂tle代码：

 #include <iostream>
 #include<cstdio>
 #include<string>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 string add(string s1,string s2)
 {
       string ans = "";
       int i,j,x,y,k=;
       for(i=s1.length()-,j=s2.length()-;i>= && j>= ;i--,j--)
       {
          x = s1[i] - '';
          y = s2[j] - '';
          ans += char((x+y+k)% + '');
          k = (x+y+k)/;
       }
       while(i>=)
       {
          x=s1[i]-'';
          ans += char ((x+k)% + '');
          k = (x+k)/;
          i--;
       }
       while(j>=)
       {
          y=s2[j]-'';
          ans += char((y+k)% + '');
          k = (y+k)/;
          j--;
       }
       if(k>)
       ans += '';
       //ans.reverse();
       reverse(ans.begin(),ans.end());
       return ans;
 }
 string mul(string s1,string s2)
 {
     string ans = "";
     int c = ;
     for(int i = s1.length()-; i >= ; i--)
     {
         //计算s1[i]*s2,结果保存在tem中
         string tem = "";
         int x = s1[i] - '',k = ;//乘数和初始余数
         for(int j = s2.length()-; j >= ; j--)
         {
             int y = s2[j] - '';
             int d = x*y+k;
             tem = char(d% + '') + tem;
             k = d/;
         }
         if(k)
         tem = char(k+'') + tem;
         for(int h = ; h < c; h++)
            tem = tem+'';
         c++;
         //tem计算完毕
         ans = add(ans,tem);
     }
     return ans;
 }
 struct mlt{
     string a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4,d1,d2,d3,d4;
     void out (){
         printf("%s %s %s %s\n", a1.c_str(), a2.c_str(), a3.c_str(), a4.c_str());
         printf("%s %s %s %s\n", b1.c_str(), b2.c_str(), b3.c_str(), b4.c_str());
         printf("%s %s %s %s\n", c1.c_str(), c2.c_str(), c3.c_str(), c4.c_str());
         printf("%s %s %s %s\n", d1.c_str(), d2.c_str(), d3.c_str(), d4.c_str());
     }
     mlt operator * (const mlt m) const
     {
         mlt tm;
         tm.a1 = add(add(add(mul(a1,m.a1),mul(a2,m.b1)),mul(a3,m.c1)),mul(a4,m.d1));
         tm.a2 = add(add(add(mul(a1,m.a2),mul(a2,m.b2)),mul(a3,m.c2)),mul(a4,m.d2));
         tm.a3 = add(add(add(mul(a1,m.a3),mul(a2,m.b3)),mul(a3,m.c3)),mul(a4,m.d3));
         tm.a4 = add(add(add(mul(a1,m.a4),mul(a2,m.b4)),mul(a3,m.c4)),mul(a4,m.d4));
         tm.b1 = add(add(add(mul(b1,m.a1),mul(b2,m.b1)),mul(b3,m.c1)),mul(b4,m.d1));
         tm.b2 = add(add(add(mul(b1,m.a2),mul(b2,m.b2)),mul(b3,m.c2)),mul(b4,m.d2));
         tm.b3 = add(add(add(mul(b1,m.a3),mul(b2,m.b3)),mul(b3,m.c3)),mul(b4,m.d3));
         tm.b4 = add(add(add(mul(b1,m.a4),mul(b2,m.b4)),mul(b3,m.c4)),mul(b4,m.d4));
         tm.c1 = add(add(add(mul(c1,m.a1),mul(c2,m.b1)),mul(c3,m.c1)),mul(c4,m.d1));
         tm.c2 = add(add(add(mul(c1,m.a2),mul(c2,m.b2)),mul(c3,m.c2)),mul(c4,m.d2));
         tm.c3 = add(add(add(mul(c1,m.a3),mul(c2,m.b3)),mul(c3,m.c3)),mul(c4,m.d3));
         tm.c4 = add(add(add(mul(c1,m.a4),mul(c2,m.b4)),mul(c3,m.c4)),mul(c4,m.d4));
         tm.d1 = add(add(add(mul(d1,m.a1),mul(d2,m.b1)),mul(d3,m.c1)),mul(d4,m.d1));
         tm.d2 = add(add(add(mul(d1,m.a2),mul(d2,m.b2)),mul(d3,m.c2)),mul(d4,m.d2));
         tm.d3 = add(add(add(mul(d1,m.a3),mul(d2,m.b3)),mul(d3,m.c3)),mul(d4,m.d3));
         tm.d4 = add(add(add(mul(d1,m.a4),mul(d2,m.b4)),mul(d3,m.c4)),mul(d4,m.d4));
     return tm;
     }
 };
 string qkmul(int n)
 {
     mlt flag;
     flag.a1 = flag.a2 = flag.a3 = flag.a4 = flag.b1 = flag.c2 = flag.d3 ="";
     flag.b2 = flag.b3 = flag.b4 = flag.c1 = flag.c3 = flag.c4 = flag.d1 = flag.d2 = flag.d4 = "" ;
     n = n-;
     mlt ret; ret.a1 = ret.b2 = ret.c3 = ret.d4 = "";
     ret.a2 = ret.a3 = ret.a4 = ret.b1 = ret.b3 = ret.b4 = ret.c1 = ret.c2 = ret.c4 = ret.d1 = ret.d2 = ret.d3 = "";
     while(n!=)
     {
         if(n%==)
             ret = ret*flag;
         n/=;
         flag = flag*flag;
     }
     //flag.out();
     flag = ret;
     string ss = "";
     ss = add(add(add(flag.a1,flag.a2),flag.a3),flag.a4);
     return ss;
 }

 int main()
 {
     int n;
     while(~scanf("%d",&n))
     {
         string sss = qkmul(n);
         cout<<sss<<endl;
      }
      return ;
 }

注明： 一般的斐波那契在n不是很大的情况可以直接用暴力的方法就可以解决了。