邀請賽隊員選拔賽–二分

文章目錄

• Problem Description
• Input
• Output
• Sample Input
• Sample Output
• Author

二分+平移LOLTime Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)Total Submission(s) : 46   Accepted Submission(s) : 7Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem DescriptionThis year, Lisi loves a game named LOL. Now He comes up with a sequence which only contains the letter 'L' and 'o'. The sequence may like this:

L o o L o o o L o o L o o o o L o o L o o o L o o

Lisi described the sequence like this : let a(0) be a 3-character sequence "L o o". Then a longer a(k) is obtained by taking a copy of a(k-1), then "L o ... o" which contains k+2 o's, and then another copy of a(k-1). For example:

a(0) = "L o o"
a(1) = "L o o L o o o L o o"
a(2) = "L o o L o o o L o o L o o o o L o o L o o o L o o"
a(3) = "L o o L o o o L o o L o o o o L o o L o o o L o o L o o o o o L o o L o o o L o o L o o o o L o o L o o o L o o"
...

Now Lisi may ask you a question. For a given number k, in the sequence a(k) which letter is on k-th position, 'L' or 'o'. For example, k = 2, the second letter in the sequence a(2) is 'o'. Again, k = 4, the 4-th letter in the sequence a(4) is 'L'.

InputEach line will contain one interger k (1 <= k <= 10^9). Process to end of file.OutputFor each case please output a letter each line, either 'L' or 'o'.Sample Input

1357910111213

Sample Output

LoooooLoo

Authorlisi

#include<iostream>#include<cstdio>using namespace std;long long a[30];void init(){    a[0]=3;    for(int i=1;i<=28;i++)        a[i]=a[i-1] + i+3 + a[i-1];        }int main(){    init();    int k;    while(cin>>k){        int pos = 0;        while(1){            for(int i=0;i<=28;i++)                if(a[i]>=k) { pos=i; break; }            if(k>3 && pos>=1) k -= a[pos-1];                        if(k==1){                printf("L\n");                break;            }            if(k>1 && k<=pos+3) {                printf("o\n");                break;            }            k-=pos+3;        }    }}