[POJ3096]Surprising Strings
試題描述
The?D-pairs?of a string of letters are the ordered pairs of letters that are distance D from each other. A string is?D-unique?if all of its D-pairs are different. A string is?surprising?if it is D-unique for every possible distance D.
Consider the string ZGBG. Its 0-pairs are ZG, GB, and BG. Since these three pairs are all different, ZGBG is 0-unique. Similarly, the 1-pairs of ZGBG are ZB and GG, and since these two pairs are different, ZGBG is 1-unique. Finally, the only 2-pair of ZGBG is ZG, so ZGBG is 2-unique. Thus ZGBG is surprising. (Note that the fact that ZG is both a 0-pair and a 2-pair of ZGBG is irrelevant, because 0 and 2 are different distances.)
Acknowledgement:?This problem is inspired by the "Puzzling Adventures" column in the December 2003 issue of?Scientific American.
輸入
The input consists of one or more nonempty strings of at most 79 uppercase letters, each string on a line by itself, followed by a line containing only an asterisk that signals the end of the input.
輸出
For each string of letters, output whether or not it is surprising using the exact output format shown below.
輸入示例
ZGBG
X
EE
AAB
AABA
AABB
BCBABCC
* 輸出示例
ZGBG is surprising. X is surprising. EE is surprising. AAB is surprising. AABA is surprising. AABB is NOT surprising. BCBABCC is NOT surprising.
數據規模及約定
見“輸入”
題解
n2?枚舉一下長度和字母對，然后把字母對哈希一下，看是否出現重復。打個時間戳可以節省時間。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;#define maxn 85
#define maxm 800
char S[maxn];
int has[maxm];int main() {while(scanf("%s", S + 1) == 1) {int n = strlen(S + 1);if(n == 1 && S[1] == '*') break;bool ok = 1;memset(has, 0, sizeof(has));for(int l = 1; l < n; l++) {for(int i = 1; i + l <= n; i++) {int x = (S[i] - 'A') * 26 + S[i+l] - 'A';if(has[x] == l){ ok = 0; break; }has[x] = l;}if(!ok) break;}printf("%s%s\n", S + 1, ok ? " is surprising." : " is NOT surprising.");}return 0;
}
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