Increase of HDU 1250 large numbers

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 6948 accepted submission (s): 2285

Problem descriptiona 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 maid number. inputeach line will contain an integers. process to end of file. outputfor each case, output the result in a line. sample input100 sample output420425145672990846840663646 Note: No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F (20) = 66526 has 5 digits. large addition, simple... But this question gives me the feeling that it hurts... It may not work, and it took more than an hour to make a mistake. The two arrays are not initialized... Then the scope of the question was not provided. I submitted it all over again and it was always wa. I thought it was my code error, and the scope of the question was from 1000 to 7100... Code:

 1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #include <map> 5 #include <iostream> 6 #include <string> 7 using namespace std; 8  9 void add(char *s1,char *s2,char *a)10 {11     int n1=strlen(s1);12     int n2=strlen(s2);13     int c1[3000], c2[3000];14     int i, j, k=0, num;15     memset(c1,0,sizeof(c1));16     memset(c2,0,sizeof(c2));17     for(i=0;i<n1;i++)18     c1[i]=s1[n1-i-1]-‘0‘;19     for(i=0;i<n2;i++)20     c2[i]=s2[n2-i-1]-‘0‘;21     n1=max(n1,n2);22     for(i=0;i<=n1;i++)23     {24         num=c1[i]+c2[i]+k;25         if(num>=10)26         {27             num-=10;28             k=1;29         }30         else k=0;31         c1[i]=num;32     }33     34     k=n1+2;35     36     while(c1[k]==0)37     k--;38     for(i=0;i<=k;i++)39     a[i]=c1[k-i]+‘0‘;40     a[k+1]=‘\0‘;41 42 }43 44 char f[8000][2050];45 46 main()47 {48      char a[3000], b[3000];49      int i, j, k, n;50      for(i=1;i<=4;i++)51      strcpy(f[i],"1");52      for(i=5;i<7100;i++)53      {54           add(f[i-1],f[i-2],a);55           add(f[i-3],f[i-4],b);56           add(a,b,f[i]);57      }58   59     while(scanf("%d",&n)==1)60      {61          62          printf("%s\n",f[n]);63      }64 }