FZU 1084 Three powers

Three powers

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Time Limit:1sMemory limit:32MAccepted Submit:96Total Submit:217

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Consider the set of all non-negative integer powers of 3.

 

S = { 1, 3, 9, 27, 81, ... }

 

Consider the sequence of all subsets of S ordered by the value of the sum of their elements. The question is simple: find the set at the n-th position in the sequence and print it in increasing order of its elements.

Input

Each line of input contains a number n, which is a positive integer with no more than 19 digits. The last line of input contains 0 and it should not be processed.

Output

For each line of input, output a single line displaying the n-th set as described above, in the format used in the sample output.

Sample input

171478311259009816340490

Output for sample input

{ }{ 3, 9 }{ 1, 9, 27 }{ 3, 9, 27, 6561, 19683 }{ 59049, 3486784401, 205891132094649, 717897987691852588770249 }

Original: Waterloo

 

解题：

       题目的意思是，3的次幂所组成的集合S，按从小到大排列组合成一些数，最小的为空，比如：{},{1},{3},{1，3},{9}···刚开始还是没发现规律，查了下资料发现二进制关系。如前几项组合成的数为0,1,3,4,9,10,12····相对应的项数为0,1,2,3,4,5,6··化为二进制0,1,10,11,100,101··于是可以发现二进制上为1的，S集合中相对应的项则有出现，但是项数是从0开始的，所以程序取的n要减去1。题目要求n是最多19位的数，预计3的次幂可能会很大，采用预先生成3的64以下的各次幂。

生成方法如下：（需要调用大数模板）

int main(){ int i,j; ofstream outfile("D://3的幂运算.txt"); bignum a; a.P(1,0); outfile<<"/""<<a<<"/","<<endl; for(i=1;i<64;i++) { a*=3; outfile<<"/""<<a<<"/","<<endl; }}

 

以下是本题的代码：

#include <iostream>using namespace std;//typedef unsigned __int64 i64; typedef unsigned long long i64;char a[64][50]={"1","3","9","27","81","243","729","2187","6561","19683","59049","177147","531441","1594323","4782969","14348907","43046721","129140163","387420489","1162261467","3486784401","10460353203","31381059609","94143178827","282429536481","847288609443","2541865828329","7625597484987","22876792454961","68630377364883","205891132094649","617673396283947","1853020188851841","5559060566555523","16677181699666569","50031545098999707","150094635296999121","450283905890997363","1350851717672992089","4052555153018976267","12157665459056928801","36472996377170786403","109418989131512359209","328256967394537077627","984770902183611232881","2954312706550833698643","8862938119652501095929","26588814358957503287787","79766443076872509863361","239299329230617529590083","717897987691852588770249","2153693963075557766310747","6461081889226673298932241","19383245667680019896796723","58149737003040059690390169","174449211009120179071170507","523347633027360537213511521","1570042899082081611640534563","4710128697246244834921603689","14130386091738734504764811067","42391158275216203514294433201","127173474825648610542883299603","381520424476945831628649898809","1144561273430837494885949696427"};int toTwo(i64 n,int twoBit[]){ int i=0; while (n) {twoBit[i++]=n%2;n/=2; } return i;}int main(){ i64 n=0;int i,j,length; int twoBit[100]; while (scanf("%llu",&n)&& n!=0)//while (scanf("%I64u",&n)&& n!=0) { length=toTwo(n-1,twoBit);printf("{");for(i=0;i<length;i++){if(twoBit[i]==1){printf(" %s",a[i]);for(j=i+1;j<length;j++)if (twoBit[j]==1)printf(", %s",a[j]);break;}}printf(" }/n"); } return 0;}