Secret CombinationTime
limit:2000MS
Memory Limit:262144KB
64bit IO Format:%i64d &%i6 4u SubmitStatusPracticecodeforces 496B
Description
You got a box with a combination lock. The lock has a display showing n digits. There is and buttons on the box with each button changes digits on the display. You are quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display was currently showing Number 579, then if we push the Firs T button, the display would show 680, and if after that we push the second button, the Display would show 068.
Know that the lock would open if the display is showing the smallest possible number so can be obtained by pushing th e buttons in some order. The leading zeros is ignored while comparing numbers. Now your task was to find the desired number.
Input
The first line contains a single integer n (1≤ n ≤1000)-the number of digits on the display.
The second line contains n digits-the Initial state of the display.
Output
Print a single line containing n digits-the desired state of the display containing the smallest possible n Umber.
Sample Input
Input
3
579
Output
024
Input
4
2014
Output
0142
1#include <stdio.h>2#include <string.h>3 intMain ()4 {5 intN;6 inti,j,k;7 Chara[1005];8 intb[1005];9 while(SCANF ("%d", &n)! =EOF)Ten { Oneb[0]=0; A for(i=1;i< +; i++) -b[i]=9; -scanf"%s", a); the for(i=0; i<n;i++) - { - intx='9'-a[i]+1; - for(j=1; j<n;j++) + { - intY= (a[(i+j)%n]-'0'+X)%Ten; + if(y<B[j]) A { at for(k=1; k<n;k++) - { -B[k]= (a[(i+k)%n]-'0'+X)%Ten; - } - Break; - } in Else if(y>B[j]) - { to Break; + } - } the } * for(i=0; i<n;i++) $printf"%d", B[i]);Panax Notoginsengprintf"\ n"); - } the return 0; +}View Code
Codeforces 496B Secret Combination