Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

The aim of data compression is to reduce redundancy in stored or communicated data. This increases effective data density and speeds up data transfer rates. One possible method to compress any binary message is the following:

Replace any maximal sequence of n 1's with the binary version of n whenever it shortens the length of the message.

For example, the compressed form of the data 11111111001001111111111111110011 becomes 10000010011110011. The original data is 32 bits long while the compressed data is only 17 bits long.

The drawback of this method is that sometimes the decompression process yields more than one result for the original message, making it impossible to obtain the original message. Write a program that determines if the original message can be obtained from the compressed data when the length of the original message (L), the number of 1's in the original message (N) and the compressed data are given. The original message will be no longer than 16 Kbytes and the compressed data will be no longer than 40 bits.

Input 

The input file contains several test cases. Each test case has two lines. The first line contains L and N and the second line contains the compressed data.

The last case is followed by a line containing two zeroes.

Output 

For each test case, output a line containing the case number (starting with 1) and a message `YES', `NO' or `NOT UNIQUE'. `YES' means that the original message can be obtained. `NO' means that the compressed data has been corrupted and the original message cannot be obtained. `NOT UNIQUE' means that more than one message could have been the original message. Follow the format shown in the sample output.

Sample Input 

32 26 
10000010011110011 
9 7 
1010101 
14 14 
111111 
0 0

Sample Output 

Case #1: YES 
Case #2: NOT UNIQUE 
Case #3: NO

Claimer: The data used in this problem is unofficial data prepared by Derek Kisman. So any mistake here does not imply mistake in the offcial judge data. Only Derek Kisman is responsible for the mistakes. Report mistakes to dkisman@acm.org

---

原本的壓縮方式是將連續的 1 計數後，用一個二進制數表示。

現在要解壓縮，解壓縮的時候知道原本的長度，以及原本有多少個 1。

問解壓縮的可能有多少種。

特別小心當 0110 壓縮時仍是 0110。
也就是說 01110 也會壓縮成 0110。

那麼易想 dp 方程，dp[i][length][ones] = ways

i 表示 處理到解壓縮字串的第幾個。

必須考慮所有可能將 1 去作轉換，也就是要窮舉當前的 1 可能有多少要被釋放。

但是為了不違反原本的情況，不會同時存在連續兩段被釋放。

也就是說 101101, 不可以被解釋為 5 個 1 與 5 個 1。

#include <stdio.h>
#include <string.h>
#include <map>
#include <algorithm>
using namespace std;
map<int, int> dp[50][16*1025];
//dp[i][length][ones] = ways
int main() {
    freopen("in.txt","r+t",stdin);
    freopen("out.txt","w+t",stdout);
    int L, N;
    int i, j, k;
    int cases = 0;
    char s[105];
    while(scanf("%d %d", &L, &N) == 2 && L > 0) {
        scanf("%s", s);
        int len = strlen(s);
        for(i = 0; i <= len; i++) {
            for(j = 0; j <= L; j++)
                dp[i][j].clear();
        }
        dp[0][0][0] = 1;
        for(i = 0; i < len; i++) {
            if(s[i] == '0') {//appends a zero.
                for(j = 0; j < L; j++) {
                    for(map<int, int>::iterator it = dp[i][j].begin();
                        it != dp[i][j].end(); it++) {
                        dp[i+1][j+1][it->first] += it->second;
                    }
                }
            } else {
                int ones = 0;
                for(j = i; s[j]; j++) {
                    ones = (ones<<1) + s[j]-'0';
                    if(ones > N+2)    break;
                    if(s[j+1] != '1') {
                        if(ones != 2) {//10 -> X
                            for(k = 0; k < L; k++) {
                                if(k + ones > L)    break;
                                for(map<int, int>::iterator it = dp[i][k].begin();
                                    it != dp[i][k].end(); it++) {
                                    if(it->first + ones > N)    break;
                                    int &val = dp[j+1][k+ones][it->first + ones];
                                    val += it->second;
                                    if(val >= 2)    val = 2;
                                }
                            }
                        }
                        if(ones == 3) {//11 -> 11
                            ones = 2;
                            for(k = 0; k < L; k++) {
                                if(k + ones > L)    break;
                                for(map<int, int>::iterator it = dp[i][k].begin();
                                    it != dp[i][k].end(); it++) {
                                    if(it->first + ones > N)    break;
                                    int &val = dp[j+1][k+ones][it->first + ones];
                                    val += it->second;
                                    if(val >= 2)    val = 2;
                                }
                            }
                            ones = 3;
                        }
                    }
                }
            }
        }
        printf("Case #%d: ", ++cases);
        int ways = dp[len][L][N];
        if(ways == 1)
            puts("YES");
        else if(ways == 0)
            puts("NO");
        else
            puts("NOT UNIQUE");
    }
    return 0;
}/*
32 27
10000010011110011
3 3
11
2 2
11
2 1
10
15 9
0110000100100
11 7
0101010011
2 1
01
9 3
00011000
15 5
100010000001000
12 4
101000010100
20 10
110100000010011010
13 7
011001001000
18 8
100101001000010000
14 4
10101000000001
0 0
*/