題干：

Problem Statement

Mr. Infinity has a string?S?consisting of digits from?1?to?9. Each time the date changes, this string changes as follows:

• Each occurrence of?2?in?S?is replaced with?22. Similarly, each?3?becomes?333,?4becomes?4444,?5?becomes?55555,?6?becomes?666666,?7?becomes?7777777,?8?becomes?88888888?and?9?becomes?999999999.?1?remains as?1.

For example, if?S?is?1324, it becomes?1333224444?the next day, and it becomes?133333333322224444444444444444?the day after next. You are interested in what the string looks like after?5×1015?days. What is the?K-th character from the left in the string after?5×1015?days?

Constraints

• S?is a string of length between?1?and?100?(inclusive).
• K?is an integer between?1?and?1018?(inclusive).
• The length of the string after?5×1015?days is at least?K.

Input

Input is given from Standard Input in the following format:

S K

Output

Print the?K-th character from the left in Mr. Infinity's string after?5×1015days.

Sample Input 1

1214 4

Sample Output 1

2

The string?S?changes as follows:

• Now:?1214
• After one day:?12214444
• After two days:?1222214444444444444444
• After three days:?12222222214444444444444444444444444444444444444444444444444444444444444444

The first five characters in the string after?5×1015?days is?12222. As?K=4, we should print the fourth character,?2.

Sample Input 2

3 157

Sample Output 2

3

The initial string is?3. The string after?5×1015?days consists only of?3.

Sample Input 3

299792458 9460730472580800

Sample Output 3

2

題目大意：

給定初始的一串?dāng)?shù)字，之后每一天每個(gè)數(shù)字會(huì)復(fù)制該數(shù)字次，比如’12223’，下一天變?yōu)椤?222222333’，如此類推，
問經(jīng)過5e15天后第k位數(shù)字是多少。

解題報(bào)告：

? ?因?yàn)槭莻€(gè)給定的數(shù)字啊！！5e15這個(gè)數(shù)字誰都承受不了啊！！讀入的字符串的大小小于100位數(shù)。所以顯然我們只需要找前綴1的個(gè)數(shù)就可以了。如果查詢的k大于前綴1的個(gè)數(shù)，那么就輸出1后面的第一個(gè)數(shù)就是了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 500 + 5; ll k,cnt; int main() {string s;cin>>s;int len = s.length();for(int i = 0; i<len; i++) {if(s[i] != '1') break;cnt++;}cin>>k;if(k <= cnt) printf("1\n");else printf("%c\n",s[cnt]);return 0 ;}

?

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