標籤:poj 數學 遞推
轉載請註明出處:http://blog.csdn.net/u012860063?viewmode=contents
題目連結:http://poj.org/problem?id=2591
Description
Set S is defined as follows:
(1) 1 is in S;
(2) If x is in S, then 2x + 1 and 3x + 1 are also in S;
(3) No other element belongs to S.
Find the N-th element of set S, if we sort the elements in S by increasing order.
Input
Input will contain several test cases; each contains a single positive integer N (1 <= N <= 10000000), which has been described above.
Output
For each test case, output the corresponding element in S.
Sample Input
100254
Sample Output
4181461
Source
POJ Monthly--2005.08.28,Static
代碼如下:
#include <iostream>using namespace std;int a[10000017];int main(){int i, two = 1, three = 1;a[1] = 1;for(i = 2; i <= 10000000; i++){a[i] = min(a[two]*2+1,a[three]*3+1);if(a[i] == a[two]*2+1)two++;if(a[i] == a[three]*3+1)three++;}int n;while(cin >> n){cout<<a[n]<<endl;}return 0;}