编程珠玑第三章
习题
1、首先是按照税率找到刚好税率最好的情况。这里用二分。
int a[100] = { 2200, 2700, ....}
double c[100] = {0, 0.14, 0.15, 0.16, ...}
首先用二分找最刚好比其小的那个数
int bs(int *a, int b, int e, int v)
{
int *begin = a + b, int *end = a + e, *mid;
if (!a || b >= e) return -1;
while (begin < end)
{
mid = begin + ((end - begin) >> 1);
if (*mid > v) end = mid;
else if (*mid < v) begin = mid + 1;
else return mid - a;
}
while (*mid > v) --mid;
return mid - a;
}
然后
int iter = bs(a, 0, 100, v);
double tax = (money - a[iter]) * c[iter+1];
2、求多项式系数
y = a[n]; for (i = n - 1; i >= 0; --i) y = x * y + a[i];
3、略过，网上见到过源代码4、日期处理
这里把所有相关的处理函数放出来。
#include <stdio.h> #include <stdlib.h> #include <string.h> int days[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; typedef struct _date { int year, month, day; }date; //看是否是闰年 int leap(int year) { return (0 == year % 4 && year % 100) || year % 400 == 0; } //看是时间是否有效 int legal(date a) { if (a.month <0 || a.month > 12) return 0; if (a.month == 2) return a.day > 0 && a.day <= (28 + leap(a.year)); return a.day >0 && a.day <= days[a.month -1]; } //比较时间 int datecmp(date a, date b) { if (a.year != b.year) return a.year - b.year; if (a.month != b.month) return a.month - b.month; return a.day - b.day; } //计算周几？ int weekday(date a) { int tm = a.month >= 3 ? (a.month - 2) : (a.month + 10); int ty = a.month >= 3 ? (a.year) : (a.year - 1); return (ty + ty/4 - ty/100 + ty/400 + (int)(2.6 * tm - 0.2) + a.day) % 7; } //时间化为天数 int date2int(date a) { int i ; int ret = a.year * 365 + (a.year - 1) / 4 - (a.year -1) / 100 + (a.year - 1) / 400; days[1] += leap(a.year); for (i = 0; i < a.month - 1; ret += days[i++]); days[1] = 28; return ret + a.day; } //天数化为时间 date int2date(int a) { date ret; int i = 0; ret.year = a/146097*400; for (a %= 146097; a >= 365+leap(ret.year); a -= 365 + leap(ret.year), ret.year++); days[i] += leap(ret.year); for (ret.month = 1; a >= days[ret.month - 1]; a -= days[ret.month - 1], ret.month++); days[1] = 28; ret.day = a + 1; return ret; } //计算距离 int dist(date a, date b) { int ad = date2int(a); int bd = date2int(b); return ad - bd > ? ad - bd : bd - ad; } //生成日历 void cal(date a) { int i, w, j, k; a.day = 1; if (a.month == 2) { k = days[a.month - 1] + leap(a.year);} else k = days[a.month - 1]; printf(" %2d月%4d年 \n", a.month, a.year); printf("日 一 二 三 四 五 六\n"); w = weekday(a); i = w % 7; while (i--) printf(" "); printf("%2d", 1); if (w % 7 == 6) printf("\n"); ++w; for (i = 1; i <= k; ++i, w = (w + 1) % 7) { if (w % 7 == 0) printf("%2d", i); else printf(" %2d", i); if (w % 7 == 6) printf("\n"); } }
第5题，解法是hash表中带hash。第一个以ic, tic 等做hash首先根据后缀找到hash_set,再把余下的字段，依次在hash_set中查找，并取出最长的。
#include<stdio.h> #include<stdlib.h> #include<string> #include<iterator> #include<iostream> #include<algorithm> #include<hash_map> #include<hash_set> using namespace std; using namespace stdext; char *p[] = {"et-ic", "al-is-tic", "s-tic", "p-tic", "-lyt-ic", "ot-ic", "an-tic", "n-tic", "c-tic", "at-ic", "h-nic", "n-ic", "m-ic", "l-lic", "b-lic", "-clic", "l-ic", "h-ic", "f-ic", "d-ic", "-bic", "a-ic", "-mac", "i-ac"}; void build_map(hash_map<string, hash_set<string> >& dict) { const int n = sizeof(p)/ sizeof(char *); for (int i = 0; i < n; ++i) { string line = p[i]; reverse(line.begin(), line.end()); int pos = line.find('-'); dict[line.substr(0, pos)].insert(line.substr(pos + 1, line.length() - pos - 1)); } } string lookup(hash_map<string, hash_set<string> >& dict, string word) { string line = word; reverse(line.begin(), line.end()); int pos = line.find('-'); string ret; hash_map<string, hash_set<string> >::iterator iter; if (dict.end() != (iter = dict.find(line.substr(0, pos)))) { hash_set<string> &h = iter->second; string temp = line.substr(pos + 1, line.length() - pos - 1); for (int j = 1; j <= temp.length(); ++j) { string c = temp.substr(0, j); if (h.find(c) != h.end() && c.length() > ret.length()) ret = c; } } ret = iter->first + "-" + ret; reverse(ret.begin(), ret.end()); return ret; } int main(void) { string sline; hash_map<string, hash_set<string> > dict; build_map(dict); while (cin >> sline) { cout << lookup(dict, sline) << endl; } return 0; }
6、这个可以用python来写，下面先看一下示例代码。模板文件/tmp/win03.domain.template
;SetupMgrTag [Unattended] [GuiUnattended] AdminPassword=${admin_password} EncryptedAdminPassword=NO OEMSkipRegional=1 [UserData] ComputerName=${computer_name}
代码import os Template = None if Template is None: t = __import__('Cheetah.Template', globals(), locals(), ['Template'], -1) Template = t.Template fp = open('/tmp/unattend.xml', 'w') tem = open('/tmp/win03.domain.template').read() domain_list = {'computer_name': 'computer', 'admin_password': 'admin', 'Use':'Use'} info = str(Template(tem, searchList=[domain_list])) fp.write(info) fp.close()如果采用C++实现，并且${}里面的代单词表明是要被替换的单词。那么处理方法如下：void Template(char *fname, char *temp_fname, map<string, string>& dict) { ofstream output(fname); ifstream temp(temp_fname); string line; while (getline(temp, line, '\n')) { int bpos = 0; string::size_type pos = 0; while (string::npos != (pos = line.find("${", bpos))) { string::size_type epos = line.find("}", pos); string t = line.substr(pos + 2, epos - pos - 2); cout << t << endl; map<string, string>::iterator iter = dict.find(t); if (dict.end() != iter) { line.replace(pos, epos - bpos + 1, iter->second); bpos += iter->second.length(); } else { bpos = epos + 1; } } output << line << "\n\r"; } temp.close(); output.close(); }
7、略去，比较扯8、2^16 = 65536个数。所以5个七段显示器肯定是够用的。
首先可以预处理一下，把0 = {0, 2, 3, 4, 5, 6}相应的8个bit置1之后的值保存到一个数组中。
char rec[10]; char temp[7]; char *get(int n) { int iter = 5; memset(temp, 0, sizeof(temp)); while (n) { temp[iter--] = rec[n % 10]; n /= 10; } while (iter>= 0) { temp[iter] = rec[0]; iter --;} return temp; }