ZOJ.2421 Recaman's Sequence【暴力打表】 2015/09/24

Recaman's Sequence

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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The Recaman's sequence is defined by a⁰ = 0 ; for m > 0, a^m = a^m-1 - m if the rsulting a^m is positive and not already in the sequence, otherwise a^m = a^m-1 + m.

The first few numbers in the Recaman's Sequence is 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9... . Given k, your task is to calculate a^k.

Input

The input consists of several test cases. Each line of the input contains an integer k where 0 <= k <= 500000. The last line contains an integer -1, which should not be processed.

Output

For each k given in the input, print one line containing a^k to the output.

Sample Input

7
10000
-1

Sample Output

20
18658

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Source: Asia 2004, Shanghai (Mainland China), Preliminary
直接打表

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;bool p[9000010];int a[500010];void init(){    memset(p,false,sizeof(p));    a[0] = 0;    for( int i = 1 ; i <= 500000 ; ++i ){        if( a[i-1] - i <= 0 || p[a[i-1]-i] )            a[i] = a[i-1] + i;        else a[i] = a[i-1] - i;        p[a[i]] = true;    }}int main(){    init();    int n;    while( ~scanf("%d",&n) ){        if( n == -1 ) break;        printf("%d\n",a[n]);    }    return 0;}