Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1031
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1031
Power Crisis
Unknown Author
Timelimit: 1
During a power crisis in New Zealand last winter (caused by a shortage of rain and hence low levels in the hydro dams), a contingency scheme was developed to turn off the power to areas of the country in a systematic, totally fair, manner. The country was divided up into N regions (Auckland was region number 1, and Wellington number 13). A number, m, would be picked `at random', and the power would first be turned off in region 1 (clearly the fairest starting point) and then in every m'th region after that, wrapping around to 1 after N, and ignoring regions already turned off. For example, if N = 17 and m = 5, power would be turned off to the regions in the order: 1,6,11,16,5,12,2,9,17,10,4,15,14,3,8,13,7.
The problem is that it is clearly fairest to turn off the region of Wellington by last(after all, that is where the Electricity headquarters are), so for a given N, the `random' number m needs to be carefully chosen so that region 13 is the last region selected.
Write a program that will read in the number of regions and then determine the smallest number m that will ensure that Wellington (region 13) can function while the rest of the country is blacked out.
Input
Input will consist of a series of lines, each line containing the number of regions N (13 ≤ N ≤ 100 ). The file will be terminated by a line consisting of a single 0.
Output
Output will consist of a series of lines, one for each line of the input. Each line will consist of the number m according to the above scheme.
| Input Sample | Output Sample |
|
17 |
7 |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int remaining(int n, int k) {
int r = 0;
for (int i = 1; i < n; i++)
r = (r + k) % i;
return r;
}
int main(){
int n, x, y, j, num, pulo;
while(1){
scanf("%d",&n);
if(n == 0) break;
y = 1;
for(;;){
if(remaining(n,y) != 11) y++;
else break;
}
printf("%d\n",y);
}
return 0;
}
Input
0
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <cstdio>
#include <deque>
#define MAXN 301
using namespace std;
int dp[MAXN][MAXN], vetor[MAXN];
void solve(int n, int m) {
deque < int> simula;
for (int i = 1; i < = n; i++) {
simula.push_back(i);
}
int cortados = 0;
while (simula.size() != 1) {
int davez = simula.front();
simula.pop_front();
if (cortados % m != 0) simula.push_back(davez);
cortados++;
}
dp[n][m] = simula.front();
}
int main() {
for (int i = 1; i < = 100; i++) {
for (int j = 1; j < MAXN; j++) {
solve(i, j);
if (dp[i][j] == 13) {
vetor[i] = j;
break;
}
}
}
int escolhido;
while (scanf("%d", &escolhido) && escolhido) {
printf("%d\n", vetor[escolhido]);
}
return 0;
}
Input
0
Output
#3 Code Example with Python Programming
Code -
Python Programming
def res(x,aux):
aux1=0
i=1
while i < x:
aux1=(aux1+aux)%i
i+=1
return aux1
while True:
x=int(input())
if x==0:break
aux=1
while res(x,aux)+2!=13:aux+=1
print(aux)
Input
0
Output