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  • POJ 1068 Parencodings

    Parencodings
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 24932   Accepted: 14695

    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

     
     
     
    解析:模拟。
     
     
     
    #include <cstdio>
    
    char s[10000000];
    int a[25];
    int n;
    int res[25];
    int rid[25];
    
    void solve()
    {
        int cnt = 0, rcnt = 0;
        for(int i = 1; i <= a[0]; ++i)
            s[cnt++] = '(';
        s[cnt] = ')';
        rid[rcnt++] = cnt;
        ++cnt;
        for(int i = 1; i < n; ++i){
            int num = a[i]-a[i-1];
            for(int j = 1; j <= num; ++j)
                s[cnt++] = '(';
            s[cnt] = ')';
            rid[rcnt++] = cnt;
            ++cnt;
        }
        int res_cnt = 0;
        for(int i = 0; i < rcnt; ++i){
            int l = 0, r = 1;
            for(int j = rid[i]-1; ; --j){
                if(s[j] == '('){
                    ++l;
                    if(l == r){
                        res[res_cnt++] = l;
                        break;
                    }
                }
                else
                    ++r;
            }
        }
        for(int i = 0; i < n-1; ++i)
            printf("%d ", res[i]);
        printf("%d
    ", res[n-1]);
    }
    
    int main()
    {
        int t;
        scanf("%d", &t);
        while(t--){
            scanf("%d", &n);
            for(int i = 0; i < n; ++i)
                scanf("%d", &a[i]);
            solve();
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/inmoonlight/p/5837971.html
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