Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6193 | Accepted: 4036 |
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
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. 题目分析:一个字符串例如:AABB,字符的两两组合为:间距为0时:AA AB BB
间距为1时:AB AB (出现相同情况,可以判断该串为 not surprise )
间距为2时: AB
代码:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <iomanip>
#include <algorithm> using namespace std; struct node
{
char a;
char b;
bool operator == (const node &x)const{
return a==x.a&&b==x.b; //重载定义==符号 判断两个结构体相等
}
}q[10000]; int main()
{
char s[100];
int i, j, k, ll;
int len;
while(scanf("%s", s) && strcmp(s, "*")!=0 ){
len = strlen(s);
if(len<=2){
printf("%s is surprising.\n", s);
}
else{
bool flag=false;
for(i=0; i<=len-2; i++)
{
int e=0;
for(j=0; j+i<len; j++)
{
q[e].a=s[j]; q[e++].b=s[j+i+1];
}
for(k=0; k<e; k++)
{
for(ll=k+1; ll<e; ll++)
{
if(q[k] == q[ll]){
flag=true; break;
}
}
}
if(flag==true){
break;
}
}
if(flag==true)
printf("%s is NOT surprising.\n", s);
else
printf("%s is surprising.\n", s);
}
}
return 0;
}