解题报告之POJ1068—Parencodings
来源:互联网 发布:知乎 鬼吹灯 编辑:程序博客网 时间:2024/06/11 19:15
Parencodings
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 11191
Accepted: 6590
Description
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
S(((()()())))
P-sequence 4 5 6666
W-sequence 1 1 1456
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.
Output
The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.
Sample Input
2
6
4 5 6 6 6 6
9
4 6 6 6 6 8 9 9 9
Sample Output
1 1 1 4 5 6
1 1 2 4 5 1 1 3 9
Source
Tehran 2001
先说下题目的意思吧
P-sequence :每个右括号之前有多少个左括号
W-sequence : 每个右括号是与之匹配的左括号之后的第几个
思路:开始觉得要用栈做,但是具体也不知道怎么应用。
于是就用白痴的模拟法做的。
先将P排列变还原成括号的形式。
然后依据括号计算W排列。
总之就是模拟人计算的过程。其实感觉这种方法比较笨,但是一时又实在想不出很巧妙的方法。它们之间应该存在某些更直接的数学关系,但是没能找到。
#include<stdio.h>main(){ int p[100]={0}; int a[100]={0}; int i,j,n,x,k; int count; while(scanf("%d",&x)!=EOF) { while(x--) { count=0; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&a[i]); if(i==0) { for(j=1;j<=a[0];j++) p[j]=1; p[j++]=-1; } if(i>0) { for(k=0;k<(a[i]-a[i-1]);k++) p[j++]=1; p[j++]=-1; } } k=0; for(i=1;p[i]!=0;i++) { if(p[i]==-1) { for(count=0,j=i;p[j]!=1;count-=p[j],j--); p[j]=0; a[k++]=count; } } for(i=0;i<k-1;i++) printf("%d ",a[i]); printf("%d\n",a[i]); } }}
- 解题报告之POJ1068—Parencodings
- Parencodings POJ1068解题报告
- POJ1068 Parencodings ACM解题报告(找规律)
- poj1068——Parencodings
- POJ1068——Parencodings
- poj1068——Parencodings
- POJ1068——Parencodings
- POJ1068解题报告
- acm-poj1068解题报告
- Parencodings poj1068
- POJ1068 Parencodings
- poj1068--Parencodings
- POJ1068-Parencodings
- POJ1068 Parencodings
- POJ1068-Parencodings
- poj1068 parencodings
- POJ1068 Parencodings
- POJ1068-Parencodings
- 三句话影响人的一生
- 人生感悟真言
- Extjs 4.2 动态更新panel内容
- Subsets II
- clearcase配置文件
- 解题报告之POJ1068—Parencodings
- 如果让自己的代码更好的让其他人理解
- Extjs4.2 grid datastore读取xml和json
- 人生经典定律
- poj 1947 Rebuilding Roads(树形DP)
- Word Ladder II 单词台阶II,求路径
- 最小树形图模版
- java中的变量
- 一些js应用