(two-dimensional array of billions or scrolling arrays) Hat's Fibonacci hdu1250

Hat ' s Fibonacci

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)

Total submission (s): 12284 Accepted Submission (s): 4124

Problem Description

A Fibonacci sequence is calculated by adding the previous and the sequence, with the first and both 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 would contain an integers. Process to end of file.

Output

For each case, the output of the result in a line.

Sample Input

100

Sample Output

4203968145672990846840663646

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

The string will time out, and the following is the timeout code.

#include <iostream>#include<string>using namespacestd;stringAddstringAstringb) {    intlen1=a.length (); intLen2=b.length (); inti; if(len1>len2) {         for(i=1; i<=len1-len2;i++) b="0"+b; }    Else    {         for(i=1; i<=len2-len1;i++) A="0"+A; }    stringstr; intcf=0, T; Len1=a.length ();  for(i=len1-1; i>=0; i--) {T=a[i]-'0'+b[i]-'0'+CF; CF=t/Ten; T%=Ten; STR=Char(t+'0')+str; }    if(cf!=0) Str=Char(cf+'0')+str; returnstr;}stringFunintN) {    stringf[10010]; f[1]="1"; f[2]="1"; f[3]="1"; f[4]="1"; inti; stringA,b,c;  for(i=5; i<=n;i++) {a=add (f[i-1],f[i-2]); b=add (f[i-3],f[i-4]); F[i]=Add (A, b); }    returnf[n];}intMain () {intN;  while(cin>>N) {stringstr; STR=Fun (n); cout<<str<<Endl; cout<<Endl; }    return 0;}

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The correct code

method One: Using two-dimensional arrays and billions of binary. #include<cstdio>#include<iostream>#include<cstring>using namespacestd;intstr[10001][260];//must be of type int. intMain () {memset (str,0,sizeof(str)); str[1][0]=1; str[2][0]=1; str[3][0]=1; str[4][0]=1; inti,j,ans=0, C,n;  for(i=5;i<10001; i++)    {         for(j=0, c=0;j<260; j + +)//Loop to enumerate the 8-bit numbers of the J arrays. {ans=str[i-1][j]+str[i-2][j]+str[i-3][j]+str[i-4][j]+C; C=ans/100000000; STR[I][J]=ans%100000000;//each number group is stored in 8 digits and C to determine whether to carry.     }    }     while(cin>>N) {j=259;  while(!str[n][j])//the first one is 0, clearing out 0. j--; cout<<str[n][j];//The first 0 of the beginning clears the X-digit number, which may be less than 8 bits.          for(i=j-1; i>=0; i--) printf ("%08d", Str[n][i]);//each 8 digit output one set, insufficient auto part 0. printf"\ n"); }    return 0;} Method Two: Solving # include with a rolling array<iostream>#include<cstdio>#include<cstring>using namespacestd; intt[6][2555]; intMain () {intN;  while(SCANF ("%d", &n)! =EOF) {memset (T,0,sizeof(t)); t[0][0] =1; t[1][0] =1; t[2][0] =1; t[3][0] =1;  for(inti =4; I < n;i++)          {              intcarry =0; intK = i%4;  for(intj =0; J <2500; j + +)              {                  intx = t[0][J] + t[1][J] + t[2][J] + t[3][j]; T[K][J]= x +carry; Carry= T[k][j]/Ten; T[K][J]%=Ten; }          }          intK =2500;  while(T[(N-1) %4][--K] = =0);  for(inti = K;i >=0; i--) {printf ("%d", t[(N-1) %4][i]); } printf ("\ n"); }                  return 0; }

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With Java

ImportJava.math.BigInteger;ImportJava.util.Scanner; Public classMain { Public Static voidMain (string[] args) {Scanner in=NewScanner (system.in); BigInteger a[]=Newbiginteger[10001]; intN;  while(In.hasnextint ()) {n=In.nextint (); a[1]=Biginteger.one; a[2]=Biginteger.one; a[3]=Biginteger.one; a[4]=Biginteger.one;  for(inti=5;i<=10000;i++) {A[i]=Biginteger.zero; A[i]=a[i].add (a[i-1]); A[i]=a[i].add (a[i-2]); A[i]=a[i].add (a[i-3]); A[i]=a[i].add (a[i-4]);        } System.out.println (A[n]); }    }}

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(two-dimensional array of billions or scrolling arrays) Hat's Fibonacci hdu1250