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NUC_HomeWork1 -- POJ1068
A - Parencodings
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
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:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
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
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
今天wj说，这道题模拟就好，用不了多长时间，结果我想了一晚上，用递推的方法做的，不知道他是怎么想的，能A出来，很开心
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 50;void print(int ans[], int n)///打印函数，其实是在中间查错的时候写的，后来就直接用了{ printf("%d", ans[1]); for(int i = 2; i <= n; ++i) { printf(" %d", ans[i]); } puts("");}int main(){#ifdef LOCAL ///重定向第一发忘记删了，错了T_T freopen("in.txt", "r", stdin);#endif int t, n; int arr1[maxn], arr2[maxn], ans[maxn];///我的想法是，预处理arr2数组用以放在此坐标左侧有几个半括号 bool tag[maxn]; ///预处理ans数组，每次值变化，都会从“1”开始，然后就处理不是“1”的值 scanf("%d", &t); ///用tag数组标记被处理过的值， while(t--) { memset(tag, false, sizeof(tag)); scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%d", &arr1[i]); ans[1] = 1; arr2[1] = arr1[1] - 1; for(int i = 2; i <= n; ++i) ///进行一次预处理,将所有为 “1” 的情形记录 { arr2[i] = arr1[i] - arr1[i-1] - 1; arr2[i] = (arr2[i] < 0) ? 0 : arr2[i]; ans[i] = (arr1[i] > arr1[i-1]) ? 1 : 0; } int i, k; int sum = 0; for(i = 2; i <= n; ++i) { if(ans[i] == 0) { for(k = i-1; k > 0; --k) ///往前找无非两种情况可以累加 { ///没有被标记过的且arr2值为0，和没有被标记过的且arr2值不为0 if((tag[k] == false) && (arr2[k] == 0)) { sum += ans[k]; tag[k] = true; ///刚开始也DB了， 用于 “==”，查了半天 } else if((tag[k] == false) && arr2[k]) { arr2[i] = --arr2[k]; arr2[k] = 0; sum += ans[k] + 1; ///找到arr2值不为‘0’，就arr2值转移到 i 身上。到这步就到底了 tag[k] = true; //printf("%d\n", arr2[i]); break; } } ans[i] = sum; sum = 0; } } print(ans, n); } return 0;}
NUC_HomeWork1 -- POJ1068
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