CF-63C - Bulls and Cows（枚举）

C - Bulls and Cows

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

The "Bulls and Cows" game needs two people to play. The thinker thinks of a number and the guesser tries to guess it.

The thinker thinks of a four-digit number in the decimal system. All the digits in the number are different and the number may have a leading zero. It can't have more than one leading zero, because all it's digits should be different. The guesser tries to guess the number. He makes a series of guesses, trying experimental numbers and receives answers from the first person in the format "x bulls y cows". xrepresents the number of digits in the experimental number that occupy the same positions as in the sought number. y

For example, let's suppose that the thinker thought of the number 0123. Then the guessers' experimental number 1263 will receive a reply "1 bull 2 cows" (3 occupies the same positions in both numbers and 1 and 2 are present in both numbers but they occupy different positions). Also, the answer to number 8103

When the guesser is answered "4 bulls 0 cows", the game is over.

Now the guesser has already made several guesses and wants to know whether his next guess can possibly be the last one.

Input

The first input line contains an integer n (1 ≤ n ≤ 10) which represents the number of already made guesses. Then follow n lines in the form of "a_i b_i c_i", where a_i is the i-th experimental number, b_i is the number of bulls, c_i is the number of cows (1 ≤ i ≤ n, 0 ≤ b_i, c_i, b_i + c_i ≤ 4). The experimental numbers are correct, i.e., each of them contains exactly four digits, in each of them all the four digits are different, and there can be a leading zero. All the experimental numbers are different. As the guesser hasn't guessed the number yet, the answer "4 bulls 0 cows" is not present.

Output

If the input data is enough to determine the sought number, print the number with four digits on a single line. If it has less than four digits, add leading zero. If the data is not enough, print "Need more data" without the quotes. If the thinker happens to have made a mistake in his replies, print "Incorrect data" without the quotes.

Sample Input

Input

21263 1 28103 2 1

Output

Need more data

Input

21234 2 21256 0 2

Output

2134

Input

20123 1 14567 1 2

Output

Incorrect data

思路：枚举0-9999，看符合条件的数有几个，没有说明信息有误，一个，答案唯一，多个，需要更多信息

#include<iostream>#include<cstring>#include<cstdio>using namespace std;class node{ public:int a,b,c;}f[111];int n;int main(){ while(cin>>n) { for(int i=0;i<n;++i) { cin>>f[i].a>>f[i].b>>f[i].c; } int ans=0,kkans; int a,b,c,d,z,bb,cc,zz; for(int i=0;i<10000;++i) { z=i;a=z%10;z/=10;b=z%10;z/=10;c=z%10;z/=10;d=z%10; if(a==b||a==c||a==d||b==c||b==d||c==d)continue; bool yes=1; for(int j=0;j<n;++j) { bb=0;cc=0; z=f[j].a; zz=z%10; if(a==zz)bb++;if(b==zz)cc++;if(c==zz)cc++;if(d==zz)cc++; z/=10;zz=z%10; if(b==zz)bb++;if(a==zz)cc++;if(c==zz)cc++;if(d==zz)cc++; z/=10;zz=z%10; if(c==zz)bb++;if(a==zz)cc++;if(b==zz)cc++;if(d==zz)cc++; z/=10;zz=z%10; if(d==zz)bb++;if(a==zz)cc++;if(b==zz)cc++;if(c==zz)cc++; if(bb!=f[j].b||cc!=f[j].c){yes=0;break;} } if(yes){ans++;kkans=i;} } if(ans==0)cout<<"Incorrect data\n"; else if(ans==1)printf("%04d\n",kkans); else cout<<"Need more data\n"; }}