Nyoj---topic 7.33 million Saints ' Day party

Halloween Party Time limit: +Ms | Memory Limit:65535KB Difficulty:1

Describe

There is a party,xadillax on Halloween and obviously going to the fun. Because the number of people to join the fun is very large, dozens of W of the Order of magnitude bar, natural to enter the need to have tickets. Luckily, Xadillax got a real ticket! The rules of the ticket arrangement are strange:

2. Ticket number is a six-digit number made up of 0~6 (0~6 these numbers can be reused)

4. Each gate ticket number of each person can not have three consecutive identical numbers (such as 123335 is not possible)

6. Each door ticket is adjacent to the two-bit difference must be under four (≤4) (such as 016245 is not possible)

Input

The first line is an n, which represents the number of inputs
Next n rows, two digits x x, y (<= y) per line

Output

For each test, output the ticket number from X to Y. A blank line between the end of each test.

Sample input

2001001 001002001011 001012

Sample output

001001001002001011001012

Source

NBOJ-1004

Uploaded by

Do not read the sentiment

Test instructions: Enter a 6-digit number, in this 6 digits can not appear more than three duplicates, can not make the adjacent two-digit difference between the value of more than 4, there is this six digits can not appear more than 6 of the number, the problem is a pit,

Do this kneeling n times ....

I have been wrong the code: because the condition is not considered the array control, the sample can be passed, has been submitted wrong n times

#include <iostream> #include <stdio.h> #include <math.h> #include <string> #include <       String.h>using namespace Std;int A[1000010];char b[10];int a1[10];bool Zhuanhua (char b[]) {for (int i=0;i<6;i++)   Convert again to reshape {a1[i]=b[i]-' 0 ';   } int flag=1;             for (int i=0;i<6;i++)//Wrong thought {if (a1[i]==a1[i+1]&&a1[i+1]==a1[i+2]) {flag=0;        Break           } if (Fabs (A1[i+1]-a1[i]) >4) {flag=0;       Break           } if (a1[i]>6) {flag=0;       Break    }} if (flag) return true; else return false;}    int main () {int n;    int x, y;    cin>>n;        while (n--) {memset (b,0,sizeof (b));        int t=0;        for (int i=x;i<=y;i++) {a[t++]=i;            } for (int i=0;i<t;i++) {sprintf (b, "%06d", A[i]); Converts an integer into a character array to store if (Zhuanhua (b)) printf ("%06d\n", A[i]);    } printf ("\ n"); } return 0;}

The correct code:

#include <iostream> #include <stdio.h> #include <math.h> #include <string> #include <        String.h>using namespace Std;int A[1000010];char b[10];int a1[10];bool Zhuanhua (char b[]) {for (int i=0;i<6;i++)   {a1[i]=b[i]-' 0 ';   } int flag=1;           for (int i=0;i<6;i++) {if (a1[i]>6) {flag=0;       Break            }} for (int i=0;i<4;i++) {if (a1[i]==a1[i+1]&&a1[i+1]==a1[i+2]) {flag=0;        Break            }} for (int i=0;i<5;i++) {if (Fabs (A1[i+1]-a1[i]) >4) {flag=0;        Break    }} if (flag) return true; else return false;}    int main () {int n;    int x, y;    cin>>n;        while (n--) {memset (b,0,sizeof (b));        int t=0;        cin>>x>>y;        for (int i=x;i<=y;i++) {a[t++]=i; } for (int i=0;i<t;i++) {sprintf (b, "%06d"), A[i]);        if (Zhuanhua (b)) printf ("%06d\n", A[i]);    } printf ("\ n"); } return 0;}

Nyoj---topic 7.33 million Saints ' Day party