hdu 1250 高精度+类似斐波那契数列

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

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7245    Accepted Submission(s): 2368

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.

就是一个简单的高精度水题

#include <string.h>#include <stdio.h>#include <iostream>#define M 2010using namespace std;char data[10000][M+2];int main(){    int i,j,p;    memset(data,0,sizeof(data));    data[1][2010]=1;    data[4][2010]=1;    data[2][2010]=1;    data[3][2010]=1;    i=5;p=2010;    //int count=0;    while(data[i-1][5]<=1)    {        for(j=2010;j>=p;j--)        {            data[i][j]=data[i-1][j]+data[i-2][j]+data[i-3][j]+data[i-4][j];        }        for(j=2010;j>=p;j--)        {            int c=data[i][j]/10;            if(c>=1)            {                data[i][j]=data[i][j]%10;                data[i][j-1]=data[i][j-1]+c;            }        }        if(data[i][p-1]>0)            p--;        i++;        //count++;    }    //printf("%d\n",count);    int n;    while(~scanf("%d",&n))    {        int temp;        for(int i=0;i<=2010;i++)            if(data[n][i]+'0'!='0')            {                temp=i;                break;            }        for(int i=temp;i<=2010;i++)            printf("%c",data[n][i]+'0');        printf("\n");    }    return 0;}