斐波那契串

https://ac.nowcoder.com/acm/contest/911/E 

思路：dfs；类似二分一样，确定位置y是前面f[i-1]还是f[i-2]，然后dfs下去；直到i==1或2；x超过最长1e8的就一直是i-1（前面的串了），y<=1e18;所以取最长和x的最小值；

#include<algorithm>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdio>
#include<map>
#include<stack>
#include<string>
#include<bits/stdc++.h>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define sfs(i) scanf("%s",(i))
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

const int maxn=5e6+9;
const int mod=1e9+7;

inline ll read()
{
    ll f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9')
    {
        if(ss=='-')f=-1;ss=getchar();
    }
    while(ss>='0'&&ss<='9')
    {
        x=x*10+ss-'0';ss=getchar();
    }    return f*x;
}

char s1[100],s2[100];
int q;
ll f[maxn];
char dfs(ll pos,ll num)
{
    if(pos==1) return s1[num];
    if(pos==2) return s2[num];
    if(num<=f[pos-1]) return dfs(pos-1,num);
    else return dfs(pos-2,num-f[pos-1]);
}
int main()
{
    //FAST_IO;
    //freopen("input.txt","r",stdin);

    sfs(s1+1);
    sfs(s2+1);
    ll len1=strlen(s1+1);
    ll len2=strlen(s2+1);
    f[1]=len1;
    f[2]=len2;
    ll maxx=0;
    for(ll i=3;;i++)
    {
        f[i]=f[i-1]+f[i-2];
        maxx=i;
        if(f[i]>=1e18) break;
    }
    cin>>q;
    while(q--)
    {
        ll x,y;
        cin>>x>>y;
        x=min(x,maxx);
        char ans=dfs(x,y);
        cout<<ans<<endl;
    }
}