cf #587 div3 E2

題目鏈接

題目坑點 k<=$10^{18}$

---

題目大意：

有序列11212312341234512345612345671234567812345678912345678910…(無限長)

現在給你一個k，問第k個位置的數字(1~9)是多少

---

大概做法：

首先給序列分層：
第1層 1
第2層 12
…
第10層 12345678910
…

先判斷k在哪一層，然後再把對應的層取出來，然後判斷k在這一層裏面的第幾個數

#include <iostream>
#include<bits/stdc++.h>
#define read(x) scanf("%d",&x)
typedef long long ll;
using namespace std;

ll s[25], ss[25];
int q;
ll k;

ll calc2(ll x)
{
    ll sum = 0, a, re = 0;
    for (ll i = 1;;i++) {
        //cout << i <<" "<< x << endl;
        if (pow(10, i) - 1 <= x){
            if (i == 1)a = 9;
            else a *= 10;
            sum += ((1 + a) * a / 2 * i);
            re =re * 10 + (a / 10 * a * (i - 1));
            sum += re;
        }else{
            ll bian = (x - ll(pow(10, i - 1)) + 1);
            sum += ((1 + bian) * bian / 2 * i);
            for (int j = 2; j <= i; j++) {
                sum += bian * 9 * ll(pow(10, j - 2)) * (j - 1);
            }
            break;
        }
    }
    return sum;
}
int main()
{
    for(int i=1;i<=17;i++){
        s[i] = 1ll*(pow(10,i)-pow(10,i-1))*i;
        ss[i] = ss[i-1] + s[i];
        //cout << ss[i] <<" ";
    }
    cin>>q;
    while(q--){
        cin>>k;
        int l = 1, r = 1e9;
        int pos;
        while(l<=r){
            int mid = (l+r)>>1;
            if(calc2(mid)>=k) {pos=mid;r = mid-1;}
            else l = mid+1;
        }
        k -= calc2(pos-1);
        l = 0; r = 17;
        while(l <=r){
            int mid = (l+r)>>1;
            if(ss[mid]>=k) {pos=mid;r = mid-1;}
            else l = mid+1;
        }
        k -= ss[pos-1];
        //cout << k<<"\teqweqweq\t" <<pos << endl;
        ll count = k/pos;
        //cout <<"cnt:" << count << endl;
        ll num = pow(10,pos-1) + count - 1;
        ll yushu = k%pos;
        if(yushu==0){
            cout<<num%10<<endl;
        }
        else{
            k -= 1ll*count*pos;
            num++;
            cout<<(num/ll(pow(10,pos-k)))%10<<endl;
        }
    }
    return 0;

}