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
14
int main() 
{
15
int ch = 1;
16 int i;
17 int invalid;
18
while ( EOF != ch ) {
19 memset( cnt, 0, sizeof(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 such 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, 0, sizeof(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, 0, sizeof(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。