poj 1019 Number Sequence 二分

标签：poj   c++   

Description

Input

Output

Sample Input

2
8
3

Sample Output

2
2

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;

const ll maxn=40000;
ll a[maxn],b[maxn];
void init(){
    a[1]=b[1]=1;
    for(ll i=2;i<maxn;i++){
        a[i]=a[i-1]+(int)log10((double)i)+1;
        b[i]=b[i-1]+a[i];
    }
    return ;
}

ll f(ll m){
    ll i=1;
    while(b[i]<m)   i++;
    int p=m-b[i-1];
    int len=0;
    for(i=1;len<p;i++)
        len+=(int)log10((double)i)+1;
    return (i-1)/(int)pow((double)10,len-p)%10;
}

int main(){
    init();
    int T;
    cin>>T;
    while(T--){
        ll n;
        cin>>n;
        cout<<f(n)<<endl;
    }
}

poj 1019 Number Sequence 二分

标签：poj   c++   

原文地址：http://blog.csdn.net/hyccfy/article/details/41453565