People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when validating, for example, whether it's still the same person behind an online avatar.
Now given a paragraph of text sampled from someone's speech, can you find the person's most commonly used word?
Input Specification:
Each input file contains one test case. For each case, there is one line of text no more than 1048576 characters in length, terminated by a carriage return ' '. The input contains at least one alphanumerical character, i.e., one character from the set [0-9 A-Z a-z].
Output Specification:
For each test case, print in one line the most commonly occurring word in the input text, followed by a space and the number of times it has occurred in the input. If there are more than one such words, print the lexicographically smallest one. The word should be printed in all lower case. Here a "word" is defined as a continuous sequence of alphanumerical characters separated by non-alphanumerical characters or the line beginning/end.
Note that words are case insensitive.
Sample Input:
Can1: "Can a can can a can? It can!"
Sample Output:
can 5
例子是个坑，一开始认为 格式是can1:”_______” ______为要检查的内容。
其实“ne character from the set [0-9 A-Z a-z]”，所以“can1”是一种字符，有别于“can”
还有就是容易超时，后来想了一个O（n）的算法，如下：
1 #include <iostream> 2 3 #include <map> 4 5 #include <string> 6 7 using namespace std; 8 9 10 11 struct word 12 13 { 14 15 int time; 16 17 int len; 18 19 }; 20 21 22 23 map<string,word> mm; 24 25 26 27 void fun1(string ss) 28 29 { 30 31 int i=0;string tem=""; 32 33 while(i<ss.length()) 34 35 { 36 37 while(i<ss.length()) 38 39 { 40 41 if((ss[i]>='A'&&ss[i]<='Z')||(ss[i]>='a'&&ss[i]<='z')||(ss[i]>='0'&&ss[i]<='9')) 42 43 { 44 45 if(ss[i]>='A'&&ss[i]<='Z') 46 47 ss[i]=ss[i]-'A'+'a'; 48 49 tem+=ss[i]; 50 51 ++i; 52 53 } 54 55 else break; 56 57 } 58 59 ++mm[tem].time; 60 61 mm[tem].len=i; 62 63 tem=""; 64 65 while(i<ss.length()) 66 67 { 68 69 if((ss[i]>='A'&&ss[i]<='Z')||(ss[i]>='a'&&ss[i]<='z')||(ss[i]>='0'&&ss[i]<='9')) 70 71 { 72 73 break; 74 75 } 76 77 else i++; 78 79 } 80 81 } 82 83 84 85 86 87 } 88 89 90 91 void fun2(string ss) 92 93 { 94 95 int i=0;string tem=""; 96 97 while(i<ss.length()) 98 99 { 100 101 while(i<ss.length()) 102 103 { 104 105 if((ss[i]>='A'&&ss[i]<='Z')||(ss[i]>='a'&&ss[i]<='z')||(ss[i]>='0'&&ss[i]<='9')) 106 107 break; 108 109 else i++; 110 111 } 112 113 while(i<ss.length()) 114 115 { 116 117 if((ss[i]>='A'&&ss[i]<='Z')||(ss[i]>='a'&&ss[i]<='z')||(ss[i]>='0'&&ss[i]<='9')) 118 119 { 120 121 if(ss[i]>='A'&&ss[i]<='Z') 122 123 ss[i]=ss[i]-'A'+'a'; 124 125 tem+=ss[i]; 126 127 ++i; 128 129 } 130 131 else break; 132 133 } 134 135 ++mm[tem].time; 136 137 mm[tem].len=i; 138 139 tem=""; 140 141 } 142 143 } 144 145 146 147 int main() 148 149 { 150 151 string ss; 152 153 int i,j;int c1,c2; 154 155 while(getline(cin,ss)) 156 157 { 158 159 mm.clear(); 160 161 162 163 164 165 if((ss[0]>='A'&&ss[0]<='Z')||(ss[0]>='a'&&ss[0]<='z')||(ss[0]>='0'&&ss[0]<='9')) 166 167 fun1(ss); 168 169 else fun2(ss); 170 171 172 173 int Max=0;int Min=1048576; 174 175 string most; 176 177 map<string,word>::iterator it; 178 179 for(it=mm.begin();it!=mm.end();it++) 180 181 { if((it->second).time>Max||((it->second).time==Max&&(it->second).len<Min)) 182 183 { 184 185 Max=(it->second).time; 186 187 Min=(it->second).len; 188 189 most=it->first; 190 191 } 192 193 } 194 195 cout<<most<<" "<<Max<<endl; 196 197 } 198 199 200 201 return 0; 202 203 }