POJ1019 Number Sequence 题解
Description
题目链接
给定一串有规律的数：11212312341234512345612345671234567812345678912345678910123456789101112345678910，问从左往右数第 i 个数字是多少？
1leq i leq 2147483647
Solution
直接把这串数字分组：1,12,123,1234,12345,123456dots
暴力打表求出每一组的数字的个数，对于每一个询问，只要先扫一遍，判断出是哪一组哪一个数字即可。
Code
代码语言：javascript复制#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<cstring>
#include<iostream>
#include<queue>
#define W while
#define I inline
#define gc getchar
#define D isdigit(c=gc())
#define Cn const
#define pc(c) putchar((c))
#define int long long
using namespace std;
I int read(){int x=0,f=1;char c;W(!D) f=c^'-'?1:-1;W(x=(x<<3) (x<<1) (c&15),D);return x*f;}
I void write(int x){x<0&&(pc('-'),x=-x),x<10?(pc(x '0'),0):(write(x/10),pc(x '0'),0);}
const int N=100010,sz=35000;
int T,n,a[sz 10],sum[sz 10],pw[20];
I void init(){a[1]=sum[1]=pw[0]=1;for(int i=1;i<=11;i ) pw[i]=pw[i-1]*10;for(int i=2;i<=sz;i ) a[i]=a[i-1] log10(i) 1,sum[i]=sum[i-1] a[i];}
signed main(){
init();T=read();W(T--){
n=read();register int i=1;W(sum[i]<n) i ;--i;
int p=n-sum[i],j=0;for(i=1;j<p;i ) j =log10(i) 1;
write((i-1)/pw[j-p]),pc('n');
}return 0;
}
