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微软2014实习生及秋令营技术类职位在线测试


1.String reorder

Time Limit: 10000ms
Case Time Limit: 1000ms
Memory Limit: 256MB

 

Description

For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’ (including ‘0’, ‘9’, ‘a’, ‘z’).

Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string. The following requirements should also be met,
1. Characters in each segment should be in strictly increasing order. For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’ (basically following ASCII character order).
2. Characters in the second segment must be the same as or a subset of the first segment; and every following segment must be the same as or a subset of its previous segment.

Your program should output string “<invalid input string>” when the input contains any invalid characters (i.e., outside the '0'-'9' and 'a'-'z' range).

 

Input

Input consists of multiple cases, one case per line. Each case is one string consisting of ASCII characters.

 

Output

For each case, print exactly one line with the reordered string based on the criteria above.

 

Sample In

aabbccdd
007799aabbccddeeff113355zz
1234.89898
abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee


Sample Out

abcdabcd
013579abcdefz013579abcdefz
<invalid input string>
abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa


代码:
 1#include <cstdio>
 2#include <cstring>
 3
 4using namespace std;
 5
 6typedef  long long  Lint;
 7
 8#define  L  256
 9Lint cnt[ L ];
10Lint tot;
11
12#define  INVALID  "<invalid input string>"
13
14int main() {
15        int ch = 1;
16        int i;
17        int invalid;
18        while ( EOF != ch ) {
19                memset( cnt, 0sizeof(cnt) );
20                for ( ch = getchar(); (EOF != ch) && ('\n' != ch); ch = getchar() ) {
21                        ++cnt[ ch ];
22                        ++tot;
23                }

24                invalid = 0;
25                for ( i = 0; i < L; ++i ) {
26                        if (    (0 < cnt[ i ]) && 
27                                (!      (       ( ('0' <= i) && ('9' >= i)
28                                                ) || 
29                                                ( ('a' <= i) && ('z' >= i)
30                                                )
31                                        )
32                                )
33                        ) {
34                                invalid = 1;
35                                break;
36                        }

37                }

38                if ( invalid ) {
39                        puts( INVALID );
40                        continue;
41                }

42
43                while ( 0 < tot ) {
44                        for ( i = 0; i < L; ++i ) {
45                                if ( 0 < cnt[ i ] ) {
46                                        putchar( i );
47                                        --cnt[ i ];
48                                        --tot;
49                                }

50                        }

51                }

52                puts( "" );
53        }

54        return 0;
55}

56




2.K-th string

Time Limit: 10000ms
Case Time Limit: 1000ms
Memory Limit: 256MB

 

Description

Consider a string set that each of them consists of {0, 1} only. All strings in the set have the same number of 0s and 1s. Write a program to find and output the K-th string according to the dictionary order. If s​uch a string doesn’t exist, or the input is not valid, please output “Impossible”. For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001.


Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case.
Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000). N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed as output.


Output

For each case, print exactly one line. If the string exists, please print it, otherwise print “Impossible”.


Sample In

3
2 2 2
2 2 7
4 7 47

Sample Out

0101
Impossible
01010111011


代码:
 1#include <cstdio>
 2#include <cstring>
 3
 4using namespace std;
 5
 6#define  IMP  "Impossible"
 7
 8typedef  long long  Lint;
 9
10#define  L  35
11
12Lint c[ L ][ L ];
13char ans[ L+L ];
14
15void init() {
16        int i, j;
17        memset( c, 0sizeof(c) );
18        c[ 0 ][ 0 ] = 1;
19        for ( i = 1; i < L; ++i ) {
20                c[ i ][ 0 ] = c[ i ][ i ] = 1;
21                for ( j = 1; j < i; ++j ) {
22                        c[ i ][ j ] = c[ i-1 ][ j-1 ] + c[ i-1 ][ j ];
23                }

24        }

25}

26
27static int solve_i( int n0, int n1, int idx, int off ) {
28        if ( (0 == n0) && (0 == n1) && (0 == idx) ) {
29                return 1;
30        }

31
32        if ( ((1 > n0) || (1 > n1)) && (1 != idx) ) {
33                return 0;
34        }

35
36        if ( 1 == idx ) {
37                for ( int i = 0; i < n0; ++i ) {
38                        ans[ off + i ] = '0';
39                }

40                for ( int i = 0; i < n1; ++i ) {
41                        ans[ off + n0 + i ] = '1';
42                }

43                return 1;
44        }

45
46        int i = 0;
47        while ( c[n1+i-1][i] < idx ) {
48                idx -= c[n1+i-1][i];
49                ++i;
50        }

51        for ( int j = 0; j < n0 - i; ++j ) {
52                ans[ off + j ] = '0';
53        }

54        ans[ off + n0 - i ] = '1';
55        return solve_i( i, n1-1, idx, off + n0 - i + 1 );
56}

57
58/*
59n0=n=4, n1=m=7, idx=k=47
60
61i   n0      n1
62
630   0000    1??????     c[6][0] = 1
641   0001    ???????     c[7][1] = 7
652   001?    ???????     c[8][2] = 28
663   01??    ???????     c[9][3] = 84
674   1???    ???????     c[10][4]
68
69*/

70
71int solve( int n, int m, int k ) {
72        if ( c[n+m][n] < k ) {
73                return 0;
74        }

75        memset( ans, 0sizeof(ans) );
76        return solve_i( n, m, k, 0 );
77}

78
79int main() {
80        int tot_case;
81        int n, m, k;
82
83        init();
84
85        scanf( "%d"&tot_case );
86        while ( 0 < tot_case-- ) {
87                scanf( "%d%d%d"&n, &m, &k );
88
89                if ( solve( n, m, k ) ) {
90                        puts( ans );
91                }

92                else {
93                        puts( IMP );
94                }

95        }

96
97        return 0;
98}

99




3.Reduce inversion count

Time Limit: 10000ms
Case Time Limit: 1000ms
Memory Limit: 256MB

Description

Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array. If such a pair exists, return the new inversion count; otherwise returns the original inversion count.

Definition of Inversion: Let (A[0], A[1] ... A[n]) be a sequence of n numbers. If i < j and A[i] > A[j], then the pair (i, j) is called inversion of A.

Example:
Count(Inversion({3, 1, 2})) = Count({3, 1}, {3, 2}) = 2
InversionCountOfSwap({3, 1, 2})=>
{
  InversionCount({1, 3, 2}) = 1 <-- swapping 1 with 3, decreases inversion count by 1
  InversionCount({2, 1, 3}) = 1 <-- swapping 2 with 3, decreases inversion count by 1
  InversionCount({3, 2, 1}) = 3 <-- swapping 1 with 2 , increases inversion count by 1
}


Input

Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma.

Output

For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced.


Sample In

3,1,2
1,2,3,4,5

Sample Out

1
0


解法:
穷举两元素交换,使用修改的归并排序求逆序对数,时间复杂度O(n*n*n*logn)。



4.Most Frequent Logs
不会。










posted on 2014-04-13 19:11 coreBugZJ 阅读(3500) 评论(0)  编辑 收藏 引用 所属分类: ACMAlgorithm


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