POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 23163 Accepted: 13574

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 6666W-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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

题目大意

 给你一个数组P，pi 等于 第i个右括号前面有多少个左括号

要你求出对应的W数组，wi 等于 第i个右括号与其对应的左括号中有多少个左括号

思路： 模拟，之后根据模拟所得，计算并且输出对应的W数组

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;#define inf 0xffffff#define maxn 22int dis[maxn],ans[maxn],vis[maxn*2],n;//dis储存输入数据，ans储存输出数据，vis储存模拟void fun(){    int flag = 0,top = 0;        //进行模拟    for(int i = 0; i < n; i++){        for(int j = flag;j < dis[i]; j++) {vis[top++] = -1;}        flag = dis[i];        vis[top++] = 1;    }        /*****************************    for(int i = 0; i < top; i++)        cout<<(vis[i]>0? ')':'(');    cout<<endl;    *****************************/        flag = 0;    for(int i = 0; i < top; i++){        if(vis[i] == 1){            int dix = 1,j = i - 1,x = 1;            while(x){                x += vis[j--];                dix++;            }            ans[flag++] = dix/2;        }    }}int main(){    int t;    scanf("%d",&t);    while(t--){        cin>>n;        for(int i = 0; i < n; i++) cin>>dis[i];        fun();        //这里是个坑，要比平时多打一个空格        for(int i = 0; i < n; i++) cout<<ans[i]<<' ';        cout<<endl;    }    return 0;}