【HDU 6130 Kolakoski】& Kolakoski & 打表

Kolakoski

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 528 Accepted Submission(s): 275

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(1≤T≤5), denoting the number of test cases.

For each test case:

A single line contains a positive integer n(1≤n≤107).

Output

For each test case:

A single line contains a nonnegative integer, denoting the answer.

Sample Input

2

1

2

Sample Output

1

2

Source

2017 Multi-University Training Contest - Team 7

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题意 ： Kolakoski 序列

思路 ： 仔细观察会发现上面的 1，2 分别代表同一个序列把相同的合并后的长度，由此可根据前面的推出后面的了

1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1

1,22,11,2,1,22,1,22,11,2,11,22,1

AC代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e7 + 10;
typedef long long LL;
int a[MAX];
void init(){
a[1] = 1;
int nl = 1;
for(int i = 1; i + nl < MAX; i++){
if(a[i] == 1){
if(a[i + nl - 1] == 1) a[i + nl] = 2;
else a[i + nl] = 1;
}
if(a[i] == 2){
a[i + nl] = a[i + nl - 1];
if(a[i + nl] == 1) a[i + ++nl] = 2;
else a[i + ++nl] = 1;
}
}
}
int main()
{
init();
int T,n; scanf("%d",&T);
while(T--){
scanf("%d",&n);
printf("%d\n",a
);
}
return 0;
}

标程 & 打表：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e7 + 10;
typedef long long LL;
int a[MAX],x,o,p;
void init(){
a[1] = x = 1,a[2] = a[3] = 2,o = p = 3;
while(p < MAX){
for(int i = 1; i <= a[o]; i++) a[++p] = x;
++o,x = 3 - x;
}
}
int main()
{
init();
int T,n; scanf("%d",&T);
while(T--){
scanf("%d",&n);
printf("%d\n",a
);
}
return 0;
}