hdu 1250 Hat's Fibonacci

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3726    Accepted Submission(s): 1252

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

4203968145672990846840663646Note: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

Recommend

Ignatius.L

 

 

这道题着实花了不少时间，倒不是花在解题思路上，而是花在了调试上面。总共提交了8次，算是最失败的了。

因为我是用打表，第一次，一个数组元素存一个十进制数，MLE。

换成了一个元素存5个十进制数，因为数组开大了点，又一次MLE。

而后两次run time error就特别怨了，也是在这花了超多是间，因为根本就看不到还有一个d%

另外需要注意的就是存五个数时，第一个需要特别处理。切记

View Code

Problem : 1250 ( Hat's Fibonacci )     Judge Status : Runtime Error
(ACCESS_VIOLATION)
RunId : 6290695    Language : C++    Author : NealGavin
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int MAX=800;int f[MAX][MAX];void goon(){ int d=0;//余数 for(int i=5;i<MAX;i++) { for(int j=1;j<=f[i-1][0];j++) { f[i][j]=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j]+d; d=f[i][j]/10; f[i][j]%=10; f[i][j]%=10; } f[i][0]=f[i-1][0]; while(d) { f[i][++f[i][0]]=d%10; d/=10; } }}int main(){ //freopen("2.out","w",stdout); memset(f,0,sizeof(f)); for(int i=1;i<5;i++) { f[i][0]=f[i][1]=1; } goon(); int m; while(scanf("%d",&m)!=EOF) { for(int i=f[m][0];i>0;i--) printf("%d",f[m][i]); printf("\n"); //printf("A%d\n",f[m][0]); }}

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int MAX=800;int f[MAX][MAX];void goon(){  int d=0;//余数  for(int i=5;i<MAX;i++)  {    for(int j=1;j<=f[i-1][0];j++)    {      f[i][j]=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j]+d;      d=f[i][j]/10;      f[i][j]%=10;                                                                                                                                                                                                                                                                                                               就是这行在编译器里一直没发现哎痛苦    ——》   f[i][j]%=10;    }    f[i][0]=f[i-1][0];    while(d)    {      f[i][++f[i][0]]=d%10;      d/=10;    }  }}int main(){  //freopen("2.out","w",stdout);  memset(f,0,sizeof(f));  for(int i=1;i<5;i++)  {    f[i][0]=f[i][1]=1;  }  goon();  int m;  while(scanf("%d",&m)!=EOF)  {    for(int i=f[m][0];i>0;i--)    printf("%d",f[m][i]);    printf("\n");    //printf("A%d\n",f[m][0]);  }}

 

 

View Code

Problem : 1250 ( Hat's Fibonacci )     Judge Status : Wrong Answer
RunId : 6291203    Language : C++    Author : NealGavin
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int MAX=8001;const int mod=100000;int f[MAX][600];void goon(){ int d=0;//余数 for(int i=5;i<MAX;i++) { for(int j=1;j<=f[i-1][0];j++) { f[i][j]+=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j]+d; if(f[i][j]>=mod) { d=f[i][j]/mod; f[i][j]%=mod; } else d=0; } f[i][0]=f[i-1][0]; while(d) { f[i][++f[i][0]]=d%mod; d/=mod; } }}int main(){ //freopen("2.out","w",stdout); memset(f,0,sizeof(f)); for(int i=1;i<5;i++) { f[i][0]=f[i][1]=1; } goon(); int m; while(scanf("%d",&m)!=EOF) { for(int i=f[m][0];i>0;i--) printf("%05d",f[m][i]); printf("\n"); // printf("A%d\n",f[m][0]); }}

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int MAX=8001;const int mod=100000;int f[MAX][600];void goon(){  int d=0;//余数  for(int i=5;i<MAX;i++)  {    for(int j=1;j<=f[i-1][0];j++)    {      f[i][j]+=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j]+d;      if(f[i][j]>=mod)      {        d=f[i][j]/mod;      f[i][j]%=mod;      }      else      d=0;    }    f[i][0]=f[i-1][0];    while(d)    {      f[i][++f[i][0]]=d%mod;      d/=mod;    }  }}int main(){  //freopen("2.out","w",stdout);  memset(f,0,sizeof(f));  for(int i=1;i<5;i++)  {    f[i][0]=f[i][1]=1;  }  goon();  int m;  while(scanf("%d",&m)!=EOF)  {    for(int i=f[m][0];i>0;i--)  就是这没处理，WA                                                                                  

     printf("%05d",f[m][i]);    printf("\n");  // printf("A%d\n",f[m][0]);  }}

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int MAX=8033;const int mod=100000;int f[MAX][800];void goon(){  int d=0;//进位  for(int i=5;i<MAX;i++)  {    for(int j=1;j<=f[i-1][0];j++)    {      f[i][j]=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j]+d;      if(f[i][j]>=mod)      {        d=f[i][j]/mod;        f[i][j]%=mod;      }      else      d=0;    }    f[i][0]=f[i-1][0];    while(d)    {      f[i][++f[i][0]]=d%mod;      d/=mod;    }  }}int main(){  //freopen("2.out","w",stdout);  memset(f,0,sizeof(f));  for(int i=1;i<5;i++)  {    f[i][0]=f[i][1]=1;  }  goon();  int m;  while(scanf("%d",&m)!=EOF)  {   printf("%d",f[m][f[m][0]]);    for(int i=f[m][0]-1;i>0;i--)    printf("%05d",f[m][i]);    printf("\n");   // printf("A%d\n",f[m][0]);  }}