目录
第三章（基本数据类型）思维导图
题目1，选做：0xCAFE的各种位运算
答案代码/补：
参考答案：
题目二，必做：判断闰年，下一天，两天时差，星期几
答案代码：个人
参考答案：
题目1，选做：0xCAFE的各种位运算
假设 int n = 0xCAFE; 请用表达式完成下面操作 (拓展题：不要求每个同学都写)
(a) 测试最后 4 位中是不是最少有 3 位为 1.
(b) 逆转字节序(i.e.,使 n = 0xFECA)
(c) 旋转 4 位 (i.e., 使 n = 0xECAF)
答案代码/补：
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdbool.h>bool is_three(int a);
int feca(int a);
int ecaf(int a);int main(void) {int n = 0xCAFE;printf("%.4x\n", n);if (is_three(n)) {printf("yes\n");}else {printf("nono\n");}printf("%.4x\n", feca(n));printf("%.4x\n", ecaf(n));return 0;
}bool is_three(int a) {int mask = a & 0xf; //只看后四位//printf("%.4x\n", mask);if (mask == 0x7 || mask == 0xb || mask > 0xc) {return true;}else {return false;}
}int feca(int a) {//0xCAFEint temp1 = a & 0xff; //temp1=0x00feint temp2 = a & 0xff00; //temp2=0xca00temp1 <<= 8; //temp1=0xfe00temp2 >>= 8; //temp2=0x00careturn temp1 | temp2;
}int ecaf(int a) {//0xCAFEint temp1 = a & 0xf; //temp1=0x000eint temp2 = a & 0xfff0; //temp2=0xcaf0temp1 <<= 12; //temp1=0xe000temp2 >>= 4; //temp2=0x0cafreturn temp1 | temp2;
}参考答案：
(a)
&
int x = n & 0xF; x == 0x7 || x == 0xB || x >= 0xD; 1111, 1110, 1101, 1011, 0111 F E D B 7(b)
(n & 0xFF) << 8; 0xFE00 n >> 8; 0x00CA ((n & 0xFF) << 8) | (n >> 8);(c)
(n & 0xF) << 12; 0xE000 (n >> 4); 0x0CAF ((n & 0xF) << 12) | (n >> 4);
题目二，必做：判断闰年，下一天，两天时差，星期几
(a) 目前使用的格里高利历闰年的规则如下：
-
公元年分非4的倍数，为平年。
-
公元年分为4的倍数但非100的倍数，为闰年。
-
公元年分为100的倍数但非400的倍数，为平年。
-
公元年分为400的倍数为闰年。
请用一个表达式判断某一年是否为闰年。
(b) 输入某一天的年月日，输出下一天的年月日。
(c) 输入某两天的年月日，输出这两天的相距多少天(不考虑公元前，且第一个日期比第二个日期要早)。
(d) 已知1970年1月1日是星期四，输入之后的某一天的年月日，判断它是星期几？
答案代码：个人
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdbool.h>
/*
(d) 已知1970年1月1日是星期四，输入之后的某一天的年月日，判断它是星期几？*/bool is_leap(int t_year);
void tomorrow(int t_year, int t_mon, int t_day,bool judge);
int interval1( int t_mon, int t_day, bool judge);
int interval2( int t_mon, int t_day, bool judge);
int interval3(int t_year,int s_year);int main(void) {int t_year, t_mon, t_day;int s_year, s_mon, s_day;printf("第一个问题：insert date\n");scanf("%d%d%d", &t_year,&t_mon,&t_day);bool judge = is_leap(t_year);if (judge==true) {printf("%d is leap year.\n", t_year);}else {printf("%d is not leap year.\n", t_year);}tomorrow(t_year, t_mon, t_day, judge);printf("第二个问题：insert first date\n");scanf("%d%d%d", &t_year, &t_mon, &t_day);printf("insert second date\n");scanf("%d%d%d", &s_year, &s_mon, &s_day);judge = is_leap(t_year);bool judge1 = is_leap(s_year);int days01 = interval1( t_mon, t_day, judge);int days02 = interval2(s_mon, s_day, judge1);int days03 = interval3(t_year,s_year);int result_days = days01 + days02 + days03;printf("%d\n", days01);printf("%d\n", days02);printf("%d\n", days03);printf("%d-%d-%d between %d-%d-%d has %d\n", t_year, t_mon, t_day, s_year, s_mon, s_day, result_days);printf("第三个问题：insert one date affter 1970,01,01\n");scanf("%d%d%d", &t_year, &t_mon, &t_day);judge = is_leap(1970);judge1 = is_leap(t_year);days01 = interval1(1, 1, judge);days02 = interval2(t_mon, t_day, judge1);days03 = interval3(1970, t_year);result_days = days01 + days02 + days03;printf("%d\n", days01);printf("%d\n", days02);printf("%d\n", days03);printf("1970-1-1 between %d-%d-%d has %d\n", t_year, t_mon, t_day, result_days);result_days %= 7;switch (result_days) {case 0:printf("周4\n"); break;case 1:printf("周5\n"); break;case 2:printf("周6\n"); break;case 3:printf("周7\n"); break;case 4:printf("周1\n"); break;case 5:printf("周2\n"); break;case 6:printf("周3\n"); break;}return 0;
}//是否是闰年
bool is_leap(int t_year) {bool judge = false;if ((t_year % 4 == 0 && t_year % 100 != 0) || (t_year % 400 == 0)) {//printf("%d is leap year.\n", t_year);judge = true;}//else {// printf("%d is not leap year.\n", t_year);//}return judge;
}//明天
void tomorrow(int t_year, int t_mon, int t_day, bool judge) {int aff_year=t_year, aff_mon=t_mon, aff_day=t_day;int month[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };if (judge == true) {month[1] = 29;}if (month[t_mon-1] == t_day) {aff_day = 1;if (t_mon == 12) {aff_year =t_year+ 1;aff_mon = 1;}else {aff_mon +=1;}}else {aff_year = t_year;aff_mon = t_mon;aff_day += 1;}printf("tomorrow is %d %d %d\n\n", aff_year, aff_mon, aff_day);
}//第一年剩余时间
int interval1( int t_mon, int t_day, bool judge) {int days = 0;int month[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };if (judge == true) {month[1] = 29;}int i = 0; for (i = 0; i < t_mon-1; i++) {days += month[i];}if (judge==true) {days = 366 - (days + t_day);}else {days = 365 - (days + t_day);}return days;
}//最后一年度过时间
int interval2(int t_mon, int t_day, bool judge) {int days = 0;int month[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };if (judge == true) {month[1] = 29;}int i = 0;for (i = 0; i < t_mon - 1; i++) {days += month[i];}days += t_day;return days;
}//中间年份时间
int interval3(int t_year, int s_year) {int days03=0;int i = t_year + 1;for (i = t_year + 1; i <= s_year-1; i++) {bool judge = is_leap(i);if (judge == true) {days03 += 366;}else {days03 += 365;}}return days03;
}
参考答案：
(a)
bool isLeapYear(int year) {return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0); }(b)
int daysOfMonth[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };void setDaysOfFeb(int year) {if (isLeapYear(year)) {daysOfMonth[2] = 29;} else {daysOfMonth[2] = 28;} }int main(void) {int year, month, day;scanf("%d%d%d", &year, &month, &day);day++;setDaysOfFeb(year);if (day > daysOfMonth[month]) {day = 1;month++;}if (month > 12) {month = 1;year++;}printf("%4d/%.2d/%2d\n", year, month, day);return 0; }(c)
int daysOfMonth[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };int distance(int year1, int month1, int day1, int year2, int month2, int day2) {int days = 0;// 计算year1还剩多少天setDaysOfFeb(year1);days += daysOfMonth[month1] - day1;for (int i = month1 + 1; i <= 12; i++) {days += daysOfMonth[i];}// 计算中间年份的天数for (int i = year1 + 1; i < year2; i++) {days += 365;if (isLeapYear(i)) {days++;}}// 计算year2过了多少天setDaysOfFeb(year2);for (int i = 1; i < month2; i++) {days += daysOfMonth[i];}days += day2;// 如果 year1 == year2, 那么多算了一整年if (year1 == year2) {days -= 365;if (isLeapYear(year1)) {days--;}}return days; }(d)
int weekday(int year, int month, int day) {int days = distance(1970, 1, 1, year, month, day);return (4 + days) % 7; }
