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google code jam Qualification Round 2012

樸素的方法水過三題

/*
2

google code jam
3

Qualification Round 2012
4

Problem a
5

*/
6

#include <stdio.h>
7

#define LM 128
9

#define LS 109
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char unmap[ LM ];
12

char words[ LS ];
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void init() {
15

int i;
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char *normal = "our language is impossible to understand there are twenty six factorial possibilities so it is okay if you want to just give up";
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char *google = "ejp mysljylc kd kxveddknmc re jsicpdrysi rbcpc ypc rtcsra dkh wyfrepkym veddknkmkrkcd de kr kd eoya kw aej tysr re ujdr lkgc jv";
18

for ( i = 0; i < LM; ++i ) {
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unmap[ i ] = i;
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}
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unmap[ 'y' ] = 'a';
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unmap[ 'e' ] = 'o';
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unmap[ 'q' ] = 'z';
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while ( (*normal) && (*google) ) {
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unmap[ *google++ ] = *normal++;
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}
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unmap[ 'z' ] = 'q';
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}
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int main() {
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int tc, cc;
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char *p;
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init();
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scanf("%d", &tc);
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gets(words);
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for ( cc = 1; cc <= tc; ++cc ) {
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gets(words);
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for ( p = words; *p; ++p ) {
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*p = unmap[ *p ];
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}
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printf("Case #%d: %s\n", cc, words);
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}
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return 0;
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}
47

/*
2

google code jam
3

Qualification Round 2012
4

Problem b
5

*/
6

#include <stdio.h>
8

#define MIN(x, y) (((x)<(y)) ? (x) : (y))
10

#define MAX(x, y) (((x)>(y)) ? (x) : (y))
11

int main() {
13

int tc, cc;
14

int n, s, p, t;
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int i;
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int ans, ans_sns, ans_s, ans_ns, res;
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scanf("%d", &tc);
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for ( cc = 1; cc <= tc; ++cc ) {
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ans_sns = ans_s = ans_ns = 0;
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scanf("%d%d%d", &n, &s, &p);
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for ( i = 0; i < n; ++i ) {
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scanf("%d", &t);
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switch( t % 3 ) {
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case 0 :
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res = t / 3;
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if ( 3 > t ) {
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if ( res >= p ) {
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++ans_ns;
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}
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}
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else {
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if ( res >= p ) {
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++ans_sns;
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}
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else if ( res + 1 >= p ) {
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++ans_s;
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}
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}
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break;
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case 1 :
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res = t / 3 + 1;
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if ( 3 > t ) {
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if ( res >= p ) {
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++ans_ns;
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}
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}
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else {
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if ( res >= p ) {
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++ans_sns;
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}
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}
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break;
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case 2 :
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res = t / 3 + 1;
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if ( res >= p ) {
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++ans_sns;
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}
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else if ( res + 1 >= p ) {
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++ans_s;
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}
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break;
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}
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}
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ans = ans_ns + MIN(ans_s, s) + ans_sns;
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printf("Case #%d: %d\n", cc, ans);
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}
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return 0;
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}
70

/*
2

google code jam
3

Qualification Round 2012
4

Problem c
5

*/
6

#include <stdio.h>
8

#include <string.h>
9

#include <stdlib.h>
10

#define L 2000009
12

void shift(char *str, int len) {
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int i;
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char tmp = str[ 0 ];
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for ( i = 1; i < len; ++i ) {
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str[ i-1 ] = str[ i ];
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}
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str[ len-1 ] = tmp;
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}
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int solve(int a, int b) {
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static char used[ L ];
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char num[ 64 ];
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int i, j, k, len, cnt, ans = 0;
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memset(used, 0, sizeof(used));
28

for ( i = a; i <= b; ++i ) {
29

if ( used[ i ] ) {
30

continue;
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}
32

itoa(i, num, 10);
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len = strlen(num);
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cnt = 0;
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for ( j = 0; j < len; ++j ) {
38

k = atoi(num);
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if ( ('0' != num[0]) && (a <= k) && (k <= b) && (! used[k]) ) {
40

++cnt;
41

used[ k ] = 1;
42

}
43

shift(num, len);
44

}
45

ans += cnt * (cnt - 1) / 2;
47

}
48

return ans;
50

}
51

int main() {
53

int tc, cc;
54

int a, b;
55

scanf("%d", &tc);
56

for ( cc = 1; cc <= tc; ++cc ) {
57

scanf("%d%d", &a, &b);
58

printf("Case #%d: %d\n", cc, solve(a, b));
59

}
60

return 0;
61

}
62

posted on 2012-04-18 23:46 coreBugZJ 閱讀(488) 評(píng)論(0) 編輯收藏引用所屬分類: ACM 、娛樂

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google code jam Qualification Round 2012