UVa 112 Tree Summing

本文介绍了一个使用LISP表达式表示二叉树的问题,目标是在树中找到一条从根到叶节点的路径,使得这些节点的值之和等于指定的整数。文章详细解释了输入格式,并提供了一段C++代码实现。

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Tree Summing

Background

LISP was one of the earliest high-level programming languages and, with FORTRAN, is one of the oldest languages currently being used. Lists, which are the fundamental data structures in LISP, can easily be adapted to represent other important data structures such as trees.

This problem deals with determining whether binary trees represented as LISP S-expressions possess a certain property.

The Problem

Given a binary tree of integers, you are to write a program that determines whether there exists a root-to-leaf path whose nodes sum to a specified integer. For example, in the tree shown below there are exactly four root-to-leaf paths. The sums of the paths are 27, 22, 26, and 18.

picture25

Binary trees are represented in the input file as LISP S-expressions having the following form.

 
empty tree ::= ()

tree ::= empty tree tex2html_wrap_inline118 (integer tree tree)

The tree diagrammed above is represented by the expression (5 (4 (11 (7 () ()) (2 () ()) ) ()) (8 (13 () ()) (4 () (1 () ()) ) ) )

Note that with this formulation all leaves of a tree are of the form (integer () () )

Since an empty tree has no root-to-leaf paths, any query as to whether a path exists whose sum is a specified integer in an empty tree must be answered negatively.

The Input

The input consists of a sequence of test cases in the form of integer/tree pairs. Each test case consists of an integer followed by one or more spaces followed by a binary tree formatted as an S-expression as described above. All binary tree S-expressions will be valid, but expressions may be spread over several lines and may contain spaces. There will be one or more test cases in an input file, and input is terminated by end-of-file.

The Output

There should be one line of output for each test case (integer/tree pair) in the input file. For each pairI,T (I represents the integer, T represents the tree) the output is the string yes if there is a root-to-leaf path in T whose sum is I and no if there is no path in T whose sum is I.

Sample Input

22 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
20 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
10 (3 
     (2 (4 () () )
        (8 () () ) )
     (1 (6 () () )
        (4 () () ) ) )
5 ()

Sample Output

yes
no
yes
no

 

这一题题意和思路都不是很难,不用建立一棵二叉树,直接进行遍历并记录节点之前包括该节点的元素之和。不过因为空格的原因,输入处理很麻烦,以前竟不知道cin.peek()和cin.ignore(),还有cin.putback()函数。看了https://github.com/igorbonadio/uva/tree/master/112上的代码才知道这回事,比直接取字符判断方便多了。这一题关键要注意空格,负号的处理,还有0()的情况。

以下是代码:

#include <iostream>
#include <string>
#include <fstream>

using namespace std;
int var;
bool ans;
void rmspace()
{
    while(cin.peek()==' '||cin.peek()=='\n')
        cin.ignore(1);
}
bool node(int sum)
{
    int v,sgn =1;
    rmspace();
    cin.ignore(1);// '('
    rmspace();
    if(cin.peek()==')')
    {
        cin.ignore(1); //")"
        return sum==var;
    }
    if(cin.peek()=='-')
    {
        cin.ignore(1);
        sgn = -1;
    }
    rmspace();
    cin>>v;
    v *= sgn;
    sum+=v;
    bool l = node(sum);
    bool r = node(sum);
    if(l&&r)
        ans = true;
    rmspace();
    cin.ignore(1);// ')'
    return false;//仅上对叶子进行判断是否等于var,非叶子用false直接跳过
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(cin>>var)
    {
        ans = 0;
        node(0);
        cout<<(ans?"yes":"no")<<endl;
    }
}
还有测试数据:

Sample Input
22 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
20 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
10 (3
(2 (4 () () )
(8 () () ) )
(1 (6 () () )
(4 () () ) ) )
5 ()
0 ()
5 (5 () ())
5 ( 5 () () )
5 (1 (3 () ()) (4 () ()))
5 (18 ( - 13 ( ) ( ))())
0 (1 ()(-2 () (1()()) ) )
2 (1 () (1 () (1 () () ) ) )
10 (5 () (5 () (5 () (5 () (4 () () ) ) ) ) )
10 (5 () (5 () (5 () (5 ( 3 () () ) (4 () () ) ) ) ) )
20 (5 () (5 () (5 () (5 () (4 () () ) ) ) ) )
Sample Output
yes
no
yes
no
no
yes
yes
yes
yes
yes
no
no
no
no
注:测试数据来自crystal_yi



### USACO 2016 January Contest Subsequences Summing to Sevens Problem Solution and Explanation In this problem from the USACO contest, one is tasked with finding the size of the largest contiguous subsequence where the sum of elements (IDs) within that subsequence is divisible by seven. The input consists of an array representing cow IDs, and the goal is to determine how many cows are part of the longest sequence meeting these criteria; if no valid sequences exist, zero should be returned. To solve this challenge efficiently without checking all possible subsequences explicitly—which would lead to poor performance—a more sophisticated approach using prefix sums modulo 7 can be applied[^1]. By maintaining a record of seen remainders when dividing cumulative totals up until each point in the list by 7 along with their earliest occurrence index, it becomes feasible to identify qualifying segments quickly whenever another instance of any remainder reappears later on during iteration through the dataset[^2]. For implementation purposes: - Initialize variables `max_length` set initially at 0 for tracking maximum length found so far. - Use dictionary or similar structure named `remainder_positions`, starting off only knowing position `-1` maps to remainder `0`. - Iterate over given numbers while updating current_sum % 7 as you go. - Check whether updated value already exists inside your tracker (`remainder_positions`). If yes, compare distance between now versus stored location against max_length variable's content—update accordingly if greater than previous best result noted down previously. - Finally add entry into mapping table linking latest encountered modulus outcome back towards its corresponding spot within enumeration process just completed successfully after loop ends normally. Below shows Python code implementing described logic effectively handling edge cases gracefully too: ```python def find_largest_subsequence_divisible_by_seven(cow_ids): max_length = 0 remainder_positions = {0: -1} current_sum = 0 for i, id_value in enumerate(cow_ids): current_sum += id_value mod_result = current_sum % 7 if mod_result not in remainder_positions: remainder_positions[mod_result] = i else: start_index = remainder_positions[mod_result] segment_size = i - start_index if segment_size > max_length: max_length = segment_size return max_length ```
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