UVa 11111 - Generalized Matrioshkas-优快云博客

本文介绍了一个关于俄罗斯套娃的通用化概念——Generalized Matrioshkas，并提供了一种验证其正确性的算法。文章详细解释了该概念的基本规则、输入输出格式及示例，并给出了一段用ANSI C实现的源代码。

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UVa 11111 - Generalized Matrioshkas

1 题目
2 思路
3 代码
4 参考

1题目

===================

Problem B - Generalized Matrioshkas

Vladimir worked for years makingmatrioshkas, those nesting dolls that certainly represent truly Russian craft. A matrioshka is a doll that may be opened in two halves, so that one finds another doll inside. Then this doll may be opened to find another one inside it. This can be repeated several times, till a final doll -that cannot be opened- is reached.

Recently, Vladimir realized that the idea of nesting dolls might be generalized to nesting toys. Indeed, he has designed toys that contain toys but in a more general sense. One of these toys may be opened in two halves and it may have more than one toy inside it. That is the new feature that Vladimir wants to introduce in his new line of toys.

Vladimir has developed a notation to describe how nesting toys should be constructed. A toy is represented with a positive integer, according to its size. More precisely: if when opening the toy represented bymwe find the toys represented byn₁,n₂,...,n_r, it must be true thatn₁+n₂+ ... +n_r<m. And if this is the case, we say that toymcontains directlythe toysn₁,n₂,...,n_r. It should be clear that toys that may be contained in any of the toysn₁,n₂,...,n_rare not considered as directly contained in the toym.

Ageneralized matrioshkais denoted with a non-empty sequence of non zero integers of the form:

a ₁ a ₂... a _N

such that toy k is represented in the sequence with two integers -k and k , with the negative one occurring in the sequence first that the positive one.

For example, the sequence

-9 -7 -22 -3 -2 -112379

represents a generalized matrioshka conformed by six toys, namely, 1 , 2 (twice), 3 , 7 and 9 . Note that toy 7 contains directly toys 2 and 3 . Note that the first copy of toy 2 occurs left from the second one and that the second copy contains directly a toy 1 . It would be wrong to understand that the first -2 and the last 2 should be paired.

On the other hand, the following sequences do not describe generalized matrioshkas:

-9 -7 -22 -3 -1 -221379

because toy2is bigger than toy1and cannot be allocated inside it.
-9 -7 -22 -3 -2 -11237 -229

because7and2may not be allocated together inside9.
-9 -7 -22 -3 -1 -232179

because there is a nesting problem within toy3.

Your problem is to write a program to help Vladimir telling good designs from bad ones.

Input

The input file contains several test cases, each one of them in a separate line. Each test case is a sequence of non zero integers, each one with an absolute value less than10⁷.

Output

Output texts for each input case are presented in the same order that input is read.

For each test case the answer must be a line of the form

:-) Matrioshka!

if the design describes a generalized matrioshka. In other case, the answer should be of the form

:-( Try again.

Sample Input

-9 -7 -2 2 -3 -2 -1 1 2 3 7 9
-9 -7 -2 2 -3 -1 -2 2 1 3 7 9
-9 -7 -2 2 -3 -1 -2 3 2 1 7 9
-100 -50 -6 6 50 100
-100 -50 -6 6 45 100
-10 -5 -2 2 5 -4 -3 3 4 10
-9 -5 -2 2 5 -4 -3 3 4 9

Sample Output

:-) Matrioshka!
:-( Try again.
:-( Try again.
:-) Matrioshka!
:-( Try again.
:-) Matrioshka!
:-( Try again.

===================

2思路

由于数据是一层一层嵌套的，所以需要两个栈，其中一个栈处理当前数据，另一个栈记录对应项的累计和，以便出栈时验证是否满足n1+n2+…+nr < m.

3代码

/*
 * Problem: 11111 - Generalized Matrioshkas
 * Lang: ANSI C
 * Time: 0.062s
 * Author: minix
 */

#include <stdio.h>
#include <string.h>
#define N 10000
int d[N], s[N], sum[N], top;

int main() {
  int i, n, num, flag;
  char c;

  while (scanf ("%d", &d[0]) != EOF) {
    c = getchar();
    n = 1; top = -1; flag = 1;
    memset (sum, 0, sizeof(sum[0])*N);
    while (c != '\n') {
      scanf ("%d", &d[n]);
      c = getchar(); n++;
    }

    for (i=0; i<n; i++) {
      if (d[i] < 0) {s[++top] = d[i];}
      else {
        if (top<0 || s[top] != -1*d[i] || d[i]<=sum[top]) {flag = 0; break;}
        else {
          sum[top] = 0;
          top--;
          if (top >=0)
            sum[top] += d[i];
        }
      }
    }

    if (top== -1 && flag == 1) printf (":-) Matrioshka!\n");
    else printf (":-( Try again.\n");
  }
  return 0;
}