题意描述
思路
AC代码
代码语言：javascript
复制#include<bits/stdc .h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x )
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=2*1e5 10;
const int M=1e6 10;
const int INF=0x3f3f3f3f;
const int MOD=1e9 7;
ll f[10];
void solve(){
ll n;cin>>n;
ll cnt=0,k=1,idx=0,t=1,cpy=n;
while(1){
cnt ;
n-=k;
//求出如果最后n小于等于0，则加上k，获得该k属于cnt段的第几个数字
if(n<=0){
n =k;
break;
}
//满足的话公差d加一
if(cnt==f[idx]){
idx ;
t ;
}
k =t;
}
string s;
//暴力构造字符串
rep(i,1,cnt 1) s =to_string(i);
cout<<s[n-1]<<endl;
}
int main(){
IOS;
f[0]=9;
rep(i,1,10) f[i]=f[i-1]*10 9;
//freopen("test.in","r",stdin);
//freopen("test.out","w",stdout);
int t;cin>>t;
while(t--)
solve();
return 0;
}
