202 lines
5.7 KiB
C
202 lines
5.7 KiB
C
#define _CRT_SECURE_NO_WARNINGS
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#include <stdio.h>
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int menu() {
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char g;
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int h;
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printf("1. 查询星期\n");
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printf("2. 打印日历\n");
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printf("输入x退出\n");
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printf("请输入选项：");
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scanf(" %c", &g);
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if (g >= '1' && g <= '2') {
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h = g - '0'; // 将字符 '1' 或 '2' 转换为整数 1 或 2
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}
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else if (g == 'x' || g == 'X') {
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h = -2; // 表示退出程序
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}
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else {
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h = -1; // 表示无效输入
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}
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return h;
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}
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//------------------------------------------------------
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int main() {
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int n;
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int m[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; //一年的月份
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for (;;) { //无限循环程序
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mjq:;
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n = menu();//用n替换h的值
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if (n == -2)
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break; //退出程序
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else if (n == 1)
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{
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int year, a = 0, b, i, j;
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printf("输入年份 月份 日期\n输入-1返回上一级\n");
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scanf("%d", &year);
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if (year == -1)
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goto mjq;//返回首页
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if (year < 0 || year>9999)//检测输入合法 万年历真的只有一万年哇！
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printf("输入错误\n\n");
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else {
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if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) //判断闰年 是的话将月份中的二月换成29天
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m[2] = 29;
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scanf("%d", &m[0]);
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scanf("%d", &b);
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if (year == 1582 && m[0] == 10 && b > 4 && b < 15) //1582年10月4日至14日之间的日期这段时间在历史上被格里高利历改革所跳过
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printf("输入错误\n1582年10月4日至14日之间的日期这段时间在历史上被格里高利历改革所跳过。\n\n");
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else if (year < 0 || m[0] < 0 || m[0]>12 || b < 0 || b>31) { //检查年份、月份和日期是否在有效范围内
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printf("输入错误\n\n");
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}
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else if (m[2] == 28 && b > 28 || m[2] == 29 && b > 29)//检查日期是否超过了该月的天数 闰年29天
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printf("输入错误\n\n");
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else if ((m[0] == 1 || m[0] == 3 || m[0] == 5 || m[0] == 7 || m[0] == 8 || m[0] == 10 || m[0] == 12) && b > 31)//如果这些月份是输入值大于31天则错误
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printf("输入错误\n\n");
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else {
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j = b;
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if (year == 1582) {
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m[10] -= 10; //1582年10月4日至14日之间的日期这段时间在历史上被格里高利历改革所跳过 减去
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if (b >= 15) //调整当月的日期
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b -= 10;
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}
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if (year >= 2000)
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a = ((year - 1) * 365 + ((year - 1) / 4) - ((year - 1) / 100) + ((year - 1) / 400) + 1) % 7;//计算从公元1年到year-1年（包括year年如果它是闰年的话）之间的闰年总数 %7计算当天是星期几
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else a = (year / 100) / 4 - 2 * (year / 100) + year % 100 + (year % 100) / 4 + (13 * 12 / 5);
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for (i = 1; i < m[0]; i++) {
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b += m[i];
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}
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a += b - 1;
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a = a % 7;
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if (a == 0)
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a = 7; //如果a=0 a就是星期天
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if (year == 1582) {
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if (b >= 15) //同上
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b -= 10;
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}
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printf("%d年%d月%d日是星期%d\n", year, m[0], j, a);
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if (m[2] == 29)
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printf("%d年是闰年\n\n", year);
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else printf("%d年不是闰年\n", year);
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}
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}
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}
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else if (n == 2)
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{
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int year, a = 0, i, j, b, f;
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cxk:;
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printf("输入年份\n输入-1返回上一级\n");
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scanf("%d", &year);
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if (year == -1)
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goto cxk;
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if (year < 0 && year > 9999)
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printf("输入错误\n\n");
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else {
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printf("1.单列输出\n2.双列输出\n输入-1返回上一级\n输入需要的序号\n");
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scanf("%d", &n);
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if (n == -1)
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goto cxk;
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if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0)
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m[2] = 29;
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if (year >= 2000)
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a = ((year - 1) * 365 + ((year - 1) / 4) - ((year - 1) / 100) + ((year - 1) / 400) + 1) % 7;
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else a = ((year - 1) * 365 + ((year - 1) / 4) - ((year - 1) / 100) + ((year - 1) / 400) + 1) % 7;
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if (a == 0)
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a = 7;
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if (n == 1)
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{
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printf("======================%.4d======================\n\n\n", year); //输出查询年份表头
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for (i = 1; i <= 12; i++) {
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printf("=======================%.2d=======================\n", i); //单列输出
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printf("星期一 星期二 星期三 星期四 星期五 星期六 星期天\n");
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for (b = 1; b < a; b++) //输出每个月空着的天数
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printf(" ");
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for (j = 1; j <= m[i]; j++) {
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if (year == 1582 && i == 10) {
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if (j == 5) //将5跳到15号
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j = 15;
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}
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printf(" %.2d ", j);
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a++;
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if (a > 7) {
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printf("\n");
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a = 1;
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}
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}printf("\n");
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}
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}
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else if (n == 2) {
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int x1, x2, x3, x4;
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printf("===============================================%.4d===============================================\n\n\n", year);
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for (i = 1; i <= 12; i += 2) {
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x2 = 1, x3 = 1;
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printf("=======================%.2d======================= ", i); //双列输出
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printf("=======================%.2d=======================\n", i + 1);
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printf("星期一 星期二 星期三 星期四 星期五 星期六 星期天 ");
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printf("星期一 星期二 星期三 星期四 星期五 星期六 星期天\n");
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for (b = 1; b < a; b++)
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printf(" ");
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f = (a + m[i]) % 7;
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if (f == 0)
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f = 7;
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x1 = 1;
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for (j = 1; j <= 40; j++) {
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if (j <= m[i])
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printf(" %.2d ", j);
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else if (x3 > m[i + 1]) {
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break;
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}
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else printf(" ");
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a++;
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if (a > 7) {
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if (x1 == 1) {
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for (b = 1; b < f; b++)
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printf(" ");
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x1 = 0;
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x2 = f, x4 = f;
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}
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for (x2; x2 <= 38; x2++) {
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if (year == 1582 && (i + 1) == 10) {
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if (x3 == 5)
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x3 = 15;
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}
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if (x3 <= m[i + 1]) {
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printf(" %.2d ", x3);
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x3++;
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}
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x4++;
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if (x4 > 7) {
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printf("\n");
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a = 1;
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x2++;
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x4 = 1;
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break;
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}
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}
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}
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}
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if (year == 1582 && (i + 1) == 10) {
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m[10] -= 10;
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}
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a = ((year - 1) * 365 + ((year - 1) / 4) - ((year - 1) / 100) + ((year - 1) / 400) + 1) % 7;
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int x5;
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for (x5 = 1; x5 <= i + 1; x5++)
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a += m[x5];
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a = a % 7;
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if (a == 0)
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a = 7;
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printf("\n");
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}
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}
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else {
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printf("输入错误\n\n");
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}
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}
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}
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else { printf("\n\n"); }
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}
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} |