uva oj 340 - Master-Mind Hints
2015-11-18 21:37
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Master-Mind Hints
MasterMind is a game for two players. One of them, Designer, selects a secret code. The other, Breaker, tries to break it. A code is no more than a row of colored dots. At the beginning of a game,the players agree upon the length N that a code must have and upon the colors that may occur in a code.
In order to break the code, Breaker makes a number of guesses, each guess itself being a code. After each guess Designer gives a hint, stating to what extent the guess matches his secret code.
In this problem you will be given a secret code
![](https://uva.onlinejudge.org/external/3/340img1.gif)
and a guess
![](https://uva.onlinejudge.org/external/3/340img2.gif)
,
and are to determine the hint. A hint consists of a pair of numbers determined as follows.
A match is a pair (i,j),
![](https://uva.onlinejudge.org/external/3/340img3.gif)
and
![](https://uva.onlinejudge.org/external/3/340img4.gif)
,
such that
![](https://uva.onlinejudge.org/external/3/340img5.gif)
. Match (i,j) is called strong when i =j, and is called weak otherwise.
Two matches (i,j) and (p,q) are called independent when i = p if and only if j = q. A set of matches is called independent when all of its members are pairwise independent.
Designer chooses an independent set M of matches for which the total number of matches and the number of strong matches are both maximal. The hint then consists of the number of strong followed by the number
of weak matches in M. Note that these numbers are uniquely determined by the secret code and the guess. If the hint turns out to be (n,0), then the guess is identical to the secret code.
Input
The input will consist of data for a number of games. The input for each game begins with an integer specifying N (the length of the code). Following these will be the secret code, represented as N integers,which we will limit to the range 1 to 9. There will then follow an arbitrary number of guesses, each also represented as N integers, each in the range 1 to 9. Following the last guess in each game will be N zeroes; these zeroes are not to
be considered as a guess.
Following the data for the first game will appear data for the second game (if any) beginning with a new value for N. The last game in the input will be followed by a single zero (when a value for N would
normally be specified). The maximum value for N will be 1000.
Output
The output for each game should list the hints that would be generated for each guess, in order, one hint per line. Each hint should be represented as a pair of integers enclosed in parentheses and separated bya comma. The entire list of hints for each game should be prefixed by a heading indicating the game number; games are numbered sequentially starting with 1. Look at the samples below for the exact format.
Sample Input
4 1 3 5 5 1 1 2 3 4 3 3 5 6 5 5 1 6 1 3 5 1 3 5 5 0 0 0 0 10 1 2 2 2 4 5 6 6 6 9 1 2 3 4 5 6 7 8 9 1 1 1 2 2 3 3 4 4 5 5 1 2 1 3 1 5 1 6 1 9 1 2 2 5 5 5 6 6 6 7 0 0 0 0 0 0 0 0 0 0 0
Sample Output
Game 1: (1,1) (2,0) (1,2) (1,2) (4,0) Game 2: (2,4) (3,2) (5,0) (7,0)
第一行给定数组长度,接下来的第一行输入 仅含1~9 标准序列,之后的是猜测序列,与彩票/双色球类似,统计有多少数字正确 A ,有多少数字在两个序列中都出现过但位置不对 B,输入全是0时结束输入。
分析:直接统计可求得A,为了求B,对每个数字,统计二者出现的次数c1 c2,则min(c1,c2)就是该数字对B的贡献,最后要减去A的部分,代码如下;
#include<stdio.h> const int mx = 1010; int n, a[mx], b[mx]; int main(){ int ca = 1; while(scanf("%d", &n)==1, n){ printf("Game %d:\n", ca++); for(int i = 0; i < n; i++) scanf("%d", &a[i]); while(1){ int A = 0, B = 0; for(int i = 0; i < n ; i++){ scanf("%d", &b[i]); if(a[i]==b[i]) A++; } if(b[0]==0) break; for(int j = 1; j <= 9; j++) { int c1 = 0, c2 = 0; for(int i = 0 ; i < n; i++) { if(a[i]==j) c1++; if(b[i]==j) c2++; } if(c1 < c2) B+=c1; else B+=c2; } printf(" (%d,%d)\n", A, B-A); } } return 0; }
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