HDU 1250 Hat's Fibonacci（大数相加）

Posted 2020-12-05 yinbiao

传送门：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): 12952    Accepted Submission(s): 4331

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 Note: 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

 

分析：

大数相加的模板题！

 

code：

#include<bits/stdc++.h>
using namespace std;
#define max_v 10005

string add(string s1,string s2)

{

    if(s1.length()<s2.length())

    {

        string temp=s1;

        s1=s2;

        s2=temp;

    }

    int i,j;

    for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)

    {

        s1[i]=char(s1[i]+(j>=0?s2[j]-‘0‘:0));   //注意细节

        if(s1[i]-‘0‘>=10)

        {

            s1[i]=char((s1[i]-‘0‘)%10+‘0‘);

            if(i) s1[i-1]++;

            else s1=‘1‘+s1;

        }

    }

    return s1;

}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        string p[n+5];
        if(n<=4)
        {
            printf("1
");
            continue;
        }
        p[1]="1";
        p[2]="1";
        p[3]="1";
        p[4]="1";
        for(int i=5; i<=n; i++)
        {
            p[i]=add(p[i-1],add(p[i-2],add(p[i-3],p[i-4])));
        }
        cout<<p[n]<<endl;
    }
    return 0;
}