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  • HDU 3130 17多校7 Kolakoski(思维简单)

    Problem Description
    This is Kolakosiki sequence: 1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1. This sequence consists of 1 and 2, and its first term equals 1. Besides, if you see adjacent and equal terms as one group, you will get 1,22,11,2,1,22,1,22,11,2,11,22,1. Count number of terms in every group, you will get the sequence itself. Now, the sequence can be uniquely determined. Please tell HazelFan its nth element.
     
    Input
    The first line contains a positive integer T(1T5), denoting the number of test cases.
    For each test case:
    A single line contains a positive integer n(1n107).
     
    Output
    For each test case:
    A single line contains a nonnegative integer, denoting the answer.
     
    Sample Input
    2
    1
    2
     
    Sample Output
    1
    2
     
    题意:
    Kolakoski序列是一个仅由1和2组成的无限数列,是一种通过“自描述”来定义的数列[1]  。他的前几项为
    1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,1,2,1,2,2,1,1,…
    它的定义很简单,若把数列中相同的数定为一组,令a(1)=1,a(2)=2,则a(n)等于第n组数的长度。
    可以根据这个定义来推算第三项以后的数:例如由于a(2)=2,因此第2组数的长度是2,因此a(3)=2,;
    由于a(3)=2,所以第三组数的长度是2,因此a(4)=a(5)=1;由于a(4)=1,a(5)=1,所以第四组数和第五组数的长度都为1,因此a(6)=2,a(7)=1,以此类推。
    给出n求第n项的数字是多少
    题解:
    直接预处理根据性质算出序列
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<queue>
     5 #include<algorithm>
     6 #include<cmath>
     7 using namespace std;
     8 const int maxn=1e7+5;
     9 int a[maxn];
    10 int b[maxn];
    11 
    12 void pre()
    13 {
    14     int cnta=1,cntb=1;
    15     a[1]=1;b[1]=1;b[2]=2;
    16     for(cnta=2,cntb=1;cnta<maxn&&cntb<maxn;cnta++)
    17     {
    18         a[cnta]=b[cnta];
    19         if(a[cnta]==1)
    20         {
    21             if(b[cntb]==1)
    22             {
    23                 b[cntb+1]=2;
    24             }
    25             else
    26             {
    27                 b[cntb+1]=1;
    28             }
    29             cntb++;
    30         }
    31         else
    32         {
    33             if(b[cntb]==1)
    34             {
    35                 b[cntb+1]=2;
    36                 b[cntb+2]=2;
    37             }
    38             else
    39             {
    40                 b[cntb+1]=1;
    41                 b[cntb+2]=1;
    42             }
    43             cntb+=2;
    44         }
    45     }
    46 }
    47 
    48 int main()
    49 {
    50     pre();
    51     int T,n;
    52     scanf("%d",&T);
    53     while(T--)
    54     {
    55         scanf("%d",&n);
    56         printf("%d
    ",b[n]);
    57     }
    58     return 0;
    59 }
     
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  • 原文地址:https://www.cnblogs.com/Annetree/p/7366166.html
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