HDU 1250 Hat's Fibonacci

题目链接：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): 7980    Accepted Submission(s): 2602

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.

 

题意很容易读懂，用高精度加法做喽，我们一般都模拟加法都模拟一位，其实，可以用压位高精度来加速。

不压位的高精度加法。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<sstream>#include<vector>#include<map>#include<stack>#include<list>#include<set>#include<queue>#define LL long long#define lson l,m,rt<<1#define rson m+1,r,rt<<1 | 1using namespace std;const int maxn=10005,maxe=100005,inf=1<<29;int n,m;char a[maxn],b[maxn],c[maxn],d[maxn],e[maxn];int na[maxn],nb[maxn],nc[maxn],nd[maxn];void add(){    int la=strlen(a),lb=strlen(b),lab=max(la,lb),lc=strlen(c),ld=strlen(d);    int lcd=max(lc,ld),lmax=(lab,lcd);    memset(na,0,sizeof(na));    memset(nb,0,sizeof(nb));    memset(nc,0,sizeof(nc));    memset(nd,0,sizeof(nd));    for(int i=0;a[i];i++) na[la-i-1]=a[i]-'0';    for(int i=0;b[i];i++) nb[lb-i-1]=b[i]-'0';    for(int i=0;c[i];i++) nc[lc-i-1]=c[i]-'0';    for(int i=0;d[i];i++) nd[ld-i-1]=d[i]-'0';    for(int i=0;i<lmax;i++) na[i]+=nb[i]+nc[i]+nd[i],na[i+1]+=na[i]/10,na[i]%=10;    if(na[lmax]) lmax++;    for(int i=0;i<lmax;i++) e[lmax-i-1]=na[i]+'0';    e[lmax]='\0';}int main(){    while(~scanf("%d",&n))    {        strcpy(a,"1");strcpy(b,"1");         strcpy(c,"1");strcpy(d,"1");         if(n<=4) {printf("1\n");continue;}        for(int i=5;i<=n;i++)        {            add();            strcpy(a,b);strcpy(b,c);            strcpy(c,d);strcpy(d,e);        }        printf("%s\n",e);    }    return 0;}

压位高精度+打表

#include<stdio.h>int a[10000][260]={0};   //每个元素可以存储8位数字，所以2005位可以用260个数组元素存储。int main(){    int i,j,n;    a[1][0]=1;      //赋初值    a[2][0]=1;    a[3][0]=1;    a[4][0]=1;    for(i=5;i<10000;i++)    {        for(j=0;j<260;j++)            a[i][j]+=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];        for(j=0;j<260;j++)                //每八位考虑进位。            if(a[i][j]>100000000)            {                a[i][j+1]+=a[i][j]/100000000;                a[i][j]=a[i][j]%100000000;            }    }    while(scanf("%d",&n)!=EOF)    {        for(j=259;j>=0;j--)            if(a[n][j]!=0) break;      //不输出高位的0        printf("%d",a[n][j]);        for(j=j-1;j>=0;j--)            printf("%08d",a[n][j]);     //每个元素存储了八位数字，所以控制输出位数为8，左边补0        printf("\n");    }    return 0;}