poj 1068 Parencodings

模拟的题型，基本难度不大，关键读懂题意：
对于给出的原括号串，存在两种数字密码串：
1.p序列：当出现匹配括号对时，从该括号对的右括号开始往左数，直到最前面的左括号数，就是pi的值。
2.w序列：当出现匹配括号对时，包含在该括号对中的所有右括号数（包括该括号对）,就是wi的值。
题目的要求：对给出的p数字串，求出对应的s串。
串长限制均为20

提示：在处理括号序列时可以使用一个小技巧，把括号序列转化为01序列，左0右1，处理时比较方便

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

Source

#include<stdio.h>#include<string.h>int main(){    int t,m,n,i,j,k;    int a[40],b[40],c[40];    while(~scanf("%d",&t))    {        while(t--)        {            memset(b,0,sizeof(b));            scanf("%d",&m);            a[0]=0;            c[0]=0;            k=1;            for(i=1; i<=m; i++)            {                scanf("%d",&a[i]);                c[i]=a[i];                for(j=1; j<=a[i]-a[i-1]; j++)                {                    b[k++]=0;                }                b[k++]=1;                c[i]=c[i-1]+j;            }            for(i=1; i<=m; i++)            {                for(j=c[i]-1; j>=0; j--)                {                    if(b[j]==0)                    {                        b[c[i]]++;                        b[j]=1;                        break;                    }                    else                        b[c[i]]++;                }                b[c[i]]--;            }             for(i=1; i<=m; i++)                printf("%d ",(b[c[i]]+1)/2);            printf("\n");        }    }    return 0;}

这道题目竟然没调试一遍AC,主要步骤是先把运来的图像模拟出来然后找就行