Parencodings
Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:
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). 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 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: Asia 2001, Tehran (Iran)
一直想怎么从P直接求W啊？？？
想不出来 就原始做法吧。 先还原 再计算W
/* 怎么直接从p序列求出w序列 ???? */ #include <stdio.h> int main() { int t,a[21]; scanf("%d",&t); while(t--) { int p[41] = {0};/*p[1] ~ p[40] 1 为左括号 -1 为右括号*/ int n,i; scanf("%d",&n); int j = 0,pos = 1;/*j左括号数*/ for(i = 1; i <= n; i++) { scanf("%d",&a[i]); while(j < a[i]) { p[pos] = 1; j++; pos++; } p[pos++] = -1; } /*for(i = 1; i <= 2*n ; i++ ) printf("%d ",p[i]);*/ int b[21],m = 1; for(i = 2*n; i >= 1; i--) { int count = 0; if(p[i] == -1) { int j = i,sum = 0; do { if(p[j] == -1) count++; sum += p[j] + p[j - 1]; j--; } while (sum != 0); b[m++] = count; } } for(i = m - 1; i >= 1; i--) { printf("%d",b[i]); if(i > 1) putchar(' '); } printf("/n"); } return 0; }
