poj 3096(map的应用)
2014-08-11 22:52
309 查看
Surprising Strings
Description
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.
Input
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.
Output
For each string of letters, output whether or not it is surprising using the exact output format shown below.
Sample Input
Sample Output
Source
Mid-Central USA 2006
直接用map就解决了,只有两个字符hash也很简单的。
AC代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 5894 | Accepted: 3861 |
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.
Input
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.
Output
For each string of letters, output whether or not it is surprising using the exact output format shown below.
Sample Input
ZGBG X EE AAB AABA AABB BCBABCC *
Sample Output
ZGBG is surprising. X is surprising. EE is surprising. AAB is surprising. AABA is surprising. AABB is NOT surprising. BCBABCC is NOT surprising.
Source
Mid-Central USA 2006
直接用map就解决了,只有两个字符hash也很简单的。
AC代码:
#include<iostream> #include<map> #include<algorithm> #include<string> #include<string.h> using namespace std; map <string , int> m[100]; char s[90]; int judge(){ int len=strlen(s); for(int i=1;i<len-1;i++){ m[i].clear(); for(int j=0;i+j<len;j++){ char ch[3]; string str; ch[0]=s[j]; ch[1]=s[i+j]; ch[2]='\0'; str=ch; if(m[i][str]) return 0; else m[i][str]=1; } } return 1; } int main(){ while(cin>>s && s[0]!='*'){ if(judge()) cout<<s<<" is surprising."<<endl; else cout<<s<<" is NOT surprising."<<endl; } return 0; }
相关文章推荐
- pair 和 map结合应用——POJ 3096
- pair 和 map结合应用——POJ 3096
- POJ 3096 Surprising Strings(map 水)
- map的基本应用--hdu--1004&&poj--2643
- Surprising Strings poj 3096 map的运用
- POJ 3096 Surprising Strings(STL map string set vector)
- poj 2153 map简单应用
- POJ 2418 Hardwood Species(STL中map的应用)
- POJ 1002 487-3279(map应用)
- POJ 题目3096 Surprising Strings(map 水)
- poj 2503 map应用 与 C 输入问题
- Surprising Strings POJ 3096 (暴力+map)
- [ACM] POJ 3096 Surprising Strings (map的使用)
- POJ训练计划3096_Surprising Strings(STL/map)
- POJ训练计划3096_Surprising Strings(STL/map)
- [POJ 3096]Surprising Strings[map]
- POJ 3096-Surprising Strings(map-相同串)
- POJ 2418——Hardwood Species c++STL(map的应用)
- [ACM] POJ 3096 Surprising Strings (map使用)
- POJ 2503 ——map应用