CodeForces 672A Summer Camp（数字游戏，枚举）

最新推荐文章于 2021-08-26 00:26:25 发布

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本文解析了CodeForces上一道关于数字序列的简单题目，通过直接查找表的方式快速求解特定位置上的数字。

http://codeforces.com/problemset/problem/672/A

Every year, hundreds of people come to summer camps, they learn new algorithms and solve hard problems.

This is your first year at summer camp, and you are asked to solve the following problem. All integers starting with 1 are written in one line. The prefix of these line is "123456789101112131415...". Your task is to print the n-th digit of this string (digits are numbered starting with 1.

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 1000) — the position of the digit you need to print.

Output

Print the n-th digit of the line.

        Examples
      
          input
        
3

          output
        
3

          input
        
11

          output
        
0

Note

In the first sample the digit at position 3 is '3', as both integers 1 and 2 consist on one digit.

In the second sample, the digit at position 11 is '0', it belongs to the integer 10.

题意：

夏令营是学习的好机会，给定一串数字1234567891011......

输出第 n 个数字的大小。

思路：

这个题目似曾相识，就像宇神出过的一样，直接打表。

Code：

#include<stdio.h>
char ans[]= {"123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499"};
int main() {
//	for(int j=1; j<500; j++)
//		printf("%d",j);
//	printf("%d****\n",sizeof(ans));
	int n;
	while(scanf("%d",&n)!=EOF) {
		printf("%c\n",ans[n-1]);
	}
	return 0;
}