UVa/数组和字符串习题集
UVa-272.
Description:
TEX is a typesetting language developed by Donald Knuth. It takes source text together with a few
typesetting instructions and produces, one hopes, a beautiful document. Beautiful documents use “
and ” to delimit quotations, rather than the mundane " which is what is provided by most keyboards.
Keyboards typically do not have an oriented double-quote, but they do have a left-single-quote and a right-single-quote '. Check your keyboard now to locate the left-single-quote key (sometimes
called the “backquote key”) and the right-single-quote key ' (sometimes called the “apostrophe” or
just “quote”). Be careful not to confuse the left-single-quote ` with the “backslash” key. TEX lets
the user type two left-single-quotes to create a left-double-quote “ and two right-single-quotes '' to create a right-double-quote ”. Most typists, however, are accustomed to delimiting their quotations with the un-oriented double-quote ".If the source contained "To be or not to be," quoth the bard, "that is the question." then the typeset document produced by TEX would not contain the desired form: “To be or not to be,” quoth the bard, “that is the question.” In order to produce the desired form, the source file must contain the sequence:To be or not to be,'' quoth the bard, that is the question.'' You are to write a program which converts text containing double-quote (") characters into text that is identical except that double-quotes have been replaced by the two-character sequences required by TEX for delimiting quotations with oriented double-quotes. The double-quote (") characters should be replaced appropriately by either if the " opens a quotation and by '' if the " closes a quotation.
Notice that the question of nested quotations does not arise: The first " must be replaced by , the next by '', the next by , the next by '', the next by ``, the next by '', and so on.
Input:
Input will consist of several lines of text containing an even number of double-quote (") characters.
Input is ended with an end-of-file character.
Output:
The text must be output exactly as it was input except that:
• the first " in each pair is replaced by two ` characters: `` and
• the second " in each pair is replaced by two ' characters: ''.
Sample Input:
"To be or not to be," quoth the Bard, "that
is the question".
The programming contestant replied: "I must disagree.
To C' or not to C', that is The Question!"
Sample Output:
To be or not to be,'' quoth the Bard, that
is the question''.
The programming contestant replied: ``I must disagree.
To C' or not to C', that is The Question!''
Codes:
//#define LOCAL
#include <cstdio>
int main()
{
#ifdef LOCAL
freopen("E:\\Temp\\input.txt", "r", stdin);
freopen("E:\\Temp\\output.txt", "w", stdout);
#endif
int c, q = 1;
while((c=getchar()) != EOF) {
if(c == '"') {
printf("%s", q?"``":"''");
q = !q;
} else
printf("%c", c);
}
return 0;
}
UVa-10082.
Description:
A common typing error is to place the hands on the keyboard one row to the right of the correct position. So ‘Q’ is typed as ‘W’ and ‘J’ is typed as ‘K’ and
so on. You are to decode a message typed in this manner.
Input:
Input consists of several lines of text. Each line may contain digits, spaces, upper case letters (except
Q, A, Z), or punctuation shown above [except back-quote (‘)]. Keys labelled with words [Tab, BackSp,
Control, etc.] are not represented in the input.
Output:
You are to replace each letter or punction symbol by the one immediately to its left on the ‘QWERTY’
keyboard shown above. Spaces in the input should be echoed in the output.
Sample Input:
O S, GOMR YPFSU/
Sample Output:
I AM FINE TODAY.
Codes:
//#define LOCAL
#include <cstdio>
char s[] = "`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./";
int main()
{
#ifdef LOCAL
freopen("E:\\Temp\\input.txt", "r", stdin);
freopen("E:\\Temp\\output.txt", "w", stdout);
#endif
int i, c;
while((c=getchar()) != EOF) {
for(i=1; s[i]&&s[i]!=c; ++i);
if(s[i]) putchar(s[i-1]);
else putchar(c);
}
return 0;
}
UVa-401.
Description:
A regular palindrome is a string of numbers or letters that is the same forward as backward. For
example, the string “ABCDEDCBA” is a palindrome because it is the same when the string is read from
left to right as when the string is read from right to left.
A mirrored string is a string for which when each of the elements of the string is changed to its
reverse (if it has a reverse) and the string is read backwards the result is the same as the original string.
For example, the string “3AIAE” is a mirrored string because ‘A’ and ‘I’ are their own reverses, and ‘3’
and ‘E’ are each others’ reverses.
A mirrored palindrome is a string that meets the criteria of a regular palindrome and the criteria
of a mirrored string. The string “ATOYOTA” is a mirrored palindrome because if the string is read
backwards, the string is the same as the original and because if each of the characters is replaced by
its reverse and the result is read backwards, the result is the same as the original string. Of course, ‘A’,
‘T’, ‘O’, and ‘Y’ are all their own reverses.
A list of all valid characters and their reverses is as follows.
Character Reverse Character Reverse Character Reverse
A A M M Y Y
B N Z 5
C O O 1 1
D P 2 S
E 3 Q 3 E
F R 4
G S 2 5 Z
H H T T 6
I I U U 7
J L V V 8 8
K W W 9
L J X X
Note that ‘0’ (zero) and ‘O’ (the letter) are considered the same character and therefore ONLY the
letter ‘O’ is a valid character.
Input:
Input consists of strings (one per line) each of which will consist of one to twenty valid characters.
There will be no invalid characters in any of the strings. Your program should read to the end of file.
Output:
For each input string, you should print the string starting in column 1 immediately followed by exactly
one of the following strings.
STRING CRITERIA
‘ -- is not a palindrome.’ if the string is not a palindrome and is not a mirrored string
‘ -- is a regular palindrome.’ if the string is a palindrome and is not a mirrored string
‘ -- is a mirrored string.’ if the string is not a palindrome and is a mirrored string
‘ -- is a mirrored palindrome.’ if the string is a palindrome and is a mirrored string
Note that the output line is to include the ‘-’s and spacing exactly as shown in the table above and
demonstrated in the Sample Output below.
In addition, after each output line, you must print an empty line.
Sample Input:
NOTAPALINDROME
ISAPALINILAPASI
2A3MEAS
ATOYOTA
Sample Output:
NOTAPALINDROME -- is not a palindrome.
ISAPALINILAPASI -- is a regular palindrome.
2A3MEAS -- is a mirrored string.
ATOYOTA -- is a mirrored palindrome.
Codes:
//#define LOCAL
#include <cstdio>
#include <cstring>
#include <cctype>
const char* rev = "A 3 HIL JM O 2TUVWXY51SE Z 8 ";
const char* msg[] = {
"not a palindrome",
"a regular palindrome",
"a mirrored string",
"a mirrored palindrome"
};
char r(char ch) {
if(isalpha(ch)) return rev[ch-'A'];
return rev[ch-'0'+25];
}
int main()
{
#ifdef LOCAL
freopen("E:\\Temp\\input.txt", "r", stdin);
freopen("E:\\Temp\\output.txt", "w", stdout);
#endif
char s[30];
while(scanf("%s", s) == 1) {
int len = strlen(s);
int p = 1, m = 1;
for(int i=0; i<(len+1)/2; ++i) {
if(s[i] != s[len-1-i]) p = 0;
if(r(s[i]) != s[len-1-i]) m = 0;
}
printf("%s -- is %s.\n\n", s, msg[m*2+p]);
}
return 0;
}
UVa-340.
Description:
MasterMind is a game for two players. One of them, Designer, selects a secret code. The other, Breaker,
tries to break it. A code is no more than a row of colored dots. At the beginning of a game, the players
agree upon the length N that a code must have and upon the colors that may occur in a code.
In order to break the code, Breaker makes a number of guesses, each guess itself being a code. After
each guess Designer gives a hint, stating to what extent the guess matches his secret code.
In this problem you will be given a secret code s 1 ...s n and a guess g 1 ...g n , and are to determine
the hint. A hint consists of a pair of numbers determined as follows.
A match is a pair (i,j), 1 ≤ i ≤ n and 1 ≤ j ≤ n, such that s i = g j . Match (i,j) is called strong
when i = j, and is called weak otherwise. Two matches (i,j) and (p,q) are called independent when
i = p if and only if j = q. A set of matches is called independent when all of its members are pairwise
independent.
Designer chooses an independent set M of matches for which the total number of matches and the
number of strong matches are both maximal. The hint then consists of the number of strong followed
by the number of weak matches in M. Note that these numbers are uniquely determined by the secret
code and the guess. If the hint turns out to be (n,0), then the guess is identical to the secret code.
Input:
The input will consist of data for a number of games. The input for each game begins with an integer
specifying N (the length of the code). Following these will be the secret code, represented as N integers,
which we will limit to the range 1 to 9. There will then follow an arbitrary number of guesses, each
also represented as N integers, each in the range 1 to 9. Following the last guess in each game will be
N zeroes; these zeroes are not to be considered as a guess.
Following the data for the first game will appear data for the second game (if any) beginning with a
new value for N. The last game in the input will be followed by a single ‘0’ (when a value for N would
normally be specified). The maximum value for N will be 1000.
Output:
The output for each game should list the hints that would be generated for each guess, in order, one hint
per line. Each hint should be represented as a pair of integers enclosed in parentheses and separated by
a comma. The entire list of hints for each game should be prefixed by a heading indicating the game
number; games are numbered sequentially starting with 1. Look at the samples below for the exact
format.
Sample Input:
4
1 3 5 5
1 1 2 3
4 3 3 5
6 5 5 1
6 1 3 5
1 3 5 5
0 0 0 0
10
1 2 2 2 4 5 6 6 6 9
1 2 3 4 5 6 7 8 9 1
1 1 2 2 3 3 4 4 5 5
1 2 1 3 1 5 1 6 1 9
1 2 2 5 5 5 6 6 6 7
0 0 0 0 0 0 0 0 0 0
0
Sample Output:
Game 1:
(1,1)
(2,0)
(1,2)
(1,2)
(4,0)
Game 2:
(2,4)
(3,2)
(5,0)
(7,0)
Codes:
//#define LOCAL
#include <cstdio>
#define maxn 1010
int main()
{
#ifdef LOCAL
freopen("E:\\Temp\\input.txt", "r", stdin);
freopen("E:\\Temp\\output.txt", "w", stdout);
#endif
int n, a[maxn], b[maxn];
int kase = 0;
while(scanf("%d", &n)==1 && n) {
printf("Game %d:\n", ++kase);
for(int i=0; i<n; ++i) scanf("%d", &a[i]);
for(;;) {
int A = 0, B = 0;
for(int i=0; i<n; ++i) {
scanf("%d", &b[i]);
if(a[i] == b[i]) ++A;
}
if(!b[0]) break;
for(int d=1; d<=9; ++d) {
int c1 = 0, c2 = 0;
for(int i=0; i<n; ++i) {
if(a[i] == d) ++c1;
if(b[i] == d) ++c2;
}
if(c1 < c2) B += c1;
else B += c2;
}
printf(" (%d,%d)\n", A, B-A);
}
}
return 0;
}
UVa-1583.
Description:
For a positive integer N, the digit-sum of N is defined as the sum of N itself and its digits. When M
is the digitsum of N, we call N a generator of M.
For example, the digit-sum of 245 is 256 (= 245 + 2 + 4 + 5). Therefore, 245 is a generator of
256.
Not surprisingly, some numbers do not have any generators and some numbers have more than one
generator. For example, the generators of 216 are 198 and 207.
You are to write a program to find the smallest generator of the given integer.
Input:
Your program is to read from standard input. The input consists of T test cases. The number of test
cases T is given in the first line of the input. Each test case takes one line containing an integer N,
1 ≤ N ≤ 100,000.
Output:
Your program is to write to standard output. Print exactly one line for each test case. The line is to
contain a generator of N for each test case. If N has multiple generators, print the smallest. If N does
not have any generators, print ‘0’.
Sample Input:
3
216
121
2005
Sample Output:
198
0
1979
Codes:
//#define LOCAL
#include <cstdio>
#include <cstring>
#define maxn 100005
int ans[maxn];
int main()
{
#ifdef LOCAL
freopen("E:\\Temp\\input.txt", "r", stdin);
freopen("E:\\Temp\\output.txt", "w", stdout);
#endif
int T, n;
memset(ans, 0, sizeof(ans));
for(int m=1; m<maxn; ++m) {
int x = m, y = m;
while(x > 0) {
y += x%10;
x /= 10;
}
if(ans[y]==0 || m<ans[y]) ans[y] = m;
}
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
printf("%d\n", ans[n]);
}
return 0;
}
UVa-1584.
Description:
Some DNA sequences exist in circular forms as in
the following figure, which shows a circular sequence
“CGAGTCAGCT”, that is, the last symbol “T” in
“CGAGTCAGCT” is connected to the first symbol “C”. We al-
ways read a circular sequence in the clockwise direction.
Since it is not easy to store a circular sequence in a com-
puter as it is, we decided to store it as a linear sequence.
However, there can be many linear sequences that are ob-
tained from a circular sequence by cutting any place of the
circular sequence. Hence, we also decided to store the linear
sequence that is lexicographically smallest among all linear
sequences that can be obtained from a circular sequence.
Your task is to find the lexicographically smallest sequence
from a given circular sequence. For the example in the figure,
the lexicographically smallest sequence is “AGCTCGAGTC”. If there are two or more linear sequences that
are lexicographically smallest, you are to find any one of them (in fact, they are the same).
Input:
The input consists of T test cases. The number of test cases T is given on the first line of the input
file. Each test case takes one line containing a circular sequence that is written as an arbitrary linear
sequence. Since the circular sequences are DNA sequences, only four symbols, ‘A’, ‘C’, ‘G’ and ‘T’, are
allowed. Each sequence has length at least 2 and at most 100.
Output:
Print exactly one line for each test case. The line is to contain the lexicographically smallest sequence
for the test case.
Sample Input:
2
CGAGTCAGCT
CTCC
Sample Output:
AGCTCGAGTC
CCCT
Codes:
//#define LOCAL
#include <cstdio>
#include <cstring>
#define maxn 105
char s[maxn];
int less(const char* s, int p, int q) {
int n = strlen(s);
for(int i=0; i<n; ++i)
if(s[(p+i)%n] != s[(q+i)%n])
return s[(p+i)%n] < s[(q+i)%n];
return 0;
}
int main()
{
#ifdef LOCAL
freopen("E:\\Temp\\input.txt", "r", stdin);
freopen("E:\\Temp\\output.txt", "w", stdout);
#endif
int T;
scanf("%d", &T);
while(T--) {
scanf("%s", s);
int ans = 0;
int n = strlen(s);
for(int i=1; i<n; ++i)
if(less(s, i, ans)) ans = i;
for(int i=0; i<n; ++i)
putchar(s[(i+ans)%n]);
putchar('\n');
}
return 0;
}最新文章
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