学步园

/*<br /> 125874を2倍すると251748となる. これは元の数125874と同じ数を含む.<br /> 2x, 3x, 4x, 5x, 6xがxと同じ数を含むような最小の正整数xを求めよ.<br /> */<br /> #include <Windows.h><br /> #include <iostream><br /> #include <vector><br /> #include <set><br /> using namespace std;<br /> bool Problem52(unsigned N)<br /> {<br /> set<int> original;<br /> set<int> doubled;<br /> for (int i=2; i<=6; i+=1)<br /> {<br /> int tmp = N;<br /> int dtmp = tmp * i;<br /> //cout << tmp << " " << dtmp << endl;<br /> while (tmp!=0)<br /> {<br /> int n = tmp % 10;<br /> original.insert(n);<br /> tmp /= 10;<br /> }<br /> while (dtmp!=0)<br /> {<br /> int n = dtmp % 10;<br /> if (!original.count(n))<br /> return false;<br /> doubled.insert(n);<br /> dtmp /= 10;<br /> }<br /> if (original != doubled)<br /> return false;<br /> }<br /> return true;<br /> }<br /> int main()<br /> {<br /> DWORD start = GetTickCount();<br /> for (int i=123456; ;i++)<br /> {<br /> if (Problem52(i))<br /> {<br /> cout << i << endl;<br /> DWORD end = GetTickCount();<br /> cout << "the running time = " << end - start << " ms." << endl;<br /> return 0;<br /> }<br /> }<br /> return 0;<br /> } 

51开始的题目难度上升了。

#include <stdio.h><br /> #include <stdlib.h><br /> inline int check(int num) {<br /> int numz[10] = {0};<br /> int swap,i ;<br /> for (swap = num; swap != 0; swap /= 10)<br /> numz[swap % 10]++;<br /> for (i = 2 ; i <= 6 ; i++) {<br /> for (swap = num * i; swap != 0; swap /= 10)<br /> if (! (numz[swap % 10])) // pure luck<br /> return 0;<br /> }<br /> return 1;<br /> }<br /> int main() {<br /> int i;<br /> for (i = 10 ;; i++)<br /> if (check(i))<br /> break;<br /> return printf("%d/n" , i);<br /> }