hdu1250--Hat's Fibonacci

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

这个题在两个时间去做的，第一次是的代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[8001][2301];
int main()
{
    int len,sum;
    memset(dp,0,sizeof(dp));
    dp[1][2300]=dp[2][2300]=dp[3][2300]=dp[4][2300]=1;
    len=1;
    for(int i=5;i<8001;i++)
    {
        sum=0;
        for(int j=2300;j>2300-len;j--)
        {
            sum+=dp[i-1][j]+dp[i-2][j]+dp[i-3][j]+dp[i-4][j];
            dp[i][j]=sum%10;
            sum/=10;
            while(j==2300-len+1&&sum>0)
            {
                dp[i][2300-len]=sum;
                len++;
                break;
            }
        }
    }
    int n;
    while(cin>>n)
    {
        int t;
        for(t=0;t<2300;t++)
        if(dp[n][t]!=0)
        break;
        for(int j=t;j<=2300;j++)
        cout<<dp[n][j];
        cout<<endl;
    }
    return 0;
}

 

但是一直（Memory Limit Exceeded）就看了大神的代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <malloc.h>
using namespace std;
void add(char *a,char *b,char *c)
{
    int i,j,k,t,lmax,lmin,tmp;
    char *s,*pmax,*pmin;
    lmax=strlen(a);
    lmin=strlen(b);
    if(lmax<lmin)
    {
        tmp=lmax;lmax=lmin;lmin=tmp;
        pmax=b;pmin=a;
    }
    else
    {
        pmax=a;pmin=b;
    }
    s=(char*)malloc(sizeof(char)*(lmax+1));
    s[0]='0';//注意此处很容易被忽略
    for(i=lmax-1,j=lmin-1,k=lmax;j>=0;i--,j--,k--)
        s[k]=pmax[i]-'0'+pmin[j];
    for(;i>=0;i--,k--)
    s[k]=pmax[i];
    for(i=lmax;i>=0;i--)
    if(s[i]>'9')
    {
        s[i]-=10;
        s[i-1]++;
    }
    if(s[0]=='0')
    {
        for(i=0;i<=lmax;i++)
        c[i-1]=s[i];
        c[i-1]='\0';
    }
    else
    {
        for(i=0;i<=lmax;i++)
        c[i]=s[i];
        c[i]='\0';
    }
    free(s);
}
char dp[8001][2301];
int main(void)
{
    for(int i=1;i<=4;i++)
    strcpy(dp[i],"1");
    for(int i=5;i<8000;i++)
    {
        char a[2301],b[2301];
        add(dp[i-1],dp[i-2],a);
        add(dp[i-3],dp[i-4],b);
        add(a,b,dp[i]);
    }
    int n;
    while(cin>>n)
    {
        cout<<dp[n]<<endl;
    }
    return 0;
}

感觉挺复杂的，不容易明白。

还是想用int数组存，就改进了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int dp[8001][410];

int main()
{
    int sum,len;
    memset(dp,0,sizeof(dp));
    dp[1][405]=dp[2][405]=dp[3][405]=dp[4][405]=1;
    len=1;
    for(int i=5;i<8000;i++)
    {
        sum=0;
        for(int j=405;j>405-len;j--)
        {
            sum=sum+dp[i-1][j]+dp[i-2][j]+dp[i-3][j]+dp[i-4][j];
            dp[i][j]=sum%100000;
            sum/=100000;
            while(j==405-len+1&&sum>0)
            {
                dp[i][405-len]=sum;
                len++;
                break;
            }
        }
    }
    int n,i;
    while(~scanf("%d",&n))
    {
        for(i=0;i<405;i++)
        if(dp[n][i]!=0)
        break;
        printf("%d",dp[n][i]);
        for(int j=i+1;j<=405;j++)
        printf("%05d",dp[n][j]);
        printf("\n");
    }
    return 0;
}

这样改一下，感觉一下子就简单了