大数加法（SDUT“斐波那契”串）4335

题目链接：https://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Contest/contestproblem/cid/2697/pid/4335.html

具体思路：倒着相加，具体看代码

AC代码：

#include<iostream>
#include<stack>
#include<iomanip>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstring>
#include<queue>
using namespace std;
# define ll long long
# define maxn 100000+1999
# define inf 0x3f3f3f3f
int a[maxn];
int b[maxn];
int temp[maxn];
int ans[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    memset(temp,0,sizeof(temp));
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    int t1=1,t2=1;
    a[1]=1;
    b[1]=1;
    temp[1]=1;
    ans[1]=1;
    ans[2]=1;
    int p=2;
    while(p<=n+100)
    {
        int len=max(t1,t2);
        for(int i=1; i<=len; i++)
        {
            temp[i]+=a[i];
            temp[i+1]+=temp[i]/10;
            temp[i]=temp[i]%10;
        }
        if(temp[len+1]!=0)
            len++;
        for(int i=len; i>=1; i--)
        {
            ans[p+len-i+1]=temp[i];
            a[i]=b[i];
            b[i]=temp[i];
        }
        p+=len;
        t1=t2;
        t2=len;
    }
    cout<<ans[n]<<endl;
    return 0;

}