Операторы break
неуместны в дневных циклах печати. После обновления номера рабочего дня вы должны выйти из цикла:
if ((j + findStartDateInMonth(1, year)) % 7 == 0) {
record1 = j + findStartDateInMonth(1, year) + 1;
break;
}
Ваш подход можно упростить и сделать более общим. Для наглядности, вот более полная версия:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static const char *MonthNames[] = {
NULL,
"January", "February", "March", "April", "May", "June",
"July", "August", "September", "October", "November", "December",
};
/* Print the Proleptic* Gregorian calendar
(*) extended into the past before its adoption */
static void printYear(int year) {
int DaysInMonth[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
const char *DayHeadings = "Su Mo Tu We Th Fr Sa";
int d1, d2, d3, y1, day;
y1 = year - 1;
day = y1 * 1461 / 4 - y1 / 100 + y1 / 400 + 1;
if (!(year % 4) && ((year % 100) || !(year % 400)))
DaysInMonth[2] = 29;
printf("%33d\n\n", year);
for (int m = 1; m <= 12; m += 3) {
int pad = 0;
for (int i = 0; i < 3; i++) {
int len = strlen(MonthNames[m + i]);
int pad1 = len + (20 - len) / 2;
printf("%*s", pad + pad1, MonthNames[m + i]);
pad = 22 - pad1;
}
printf("\n%s %s %s\n", DayHeadings, DayHeadings, DayHeadings);
d1 = 1 - day % 7;
day += DaysInMonth[m];
d2 = 1 - day % 7;
day += DaysInMonth[m + 1];
d3 = 1 - day % 7;
day += DaysInMonth[m + 2];
for (int j = 0; j < 6; j++) {
pad = -1;
for (int i = 0; i < 7; i++, d1++) {
pad += 3;
if (d1 > 0 && d1 <= DaysInMonth[m])
pad -= printf("%*d", pad, d1);
}
pad += 1;
for (int i = 0; i < 7; i++, d2++) {
pad += 3;
if (d2 > 0 && d2 <= DaysInMonth[m + 1])
pad -= printf("%*d", pad, d2);
}
pad += 1;
for (int i = 0; i < 7; i++, d3++) {
pad += 3;
if (d3 > 0 && d3 <= DaysInMonth[m + 2])
pad -= printf("%*d", pad, d3);
}
printf("\n");
}
}
}
int main(int argc, char *argv[]) {
if (argc > 1) {
for (int i = 1; i < argc; i++) {
int year = strtol(argv[i], NULL, 0);
printYear(year);
}
} else {
printYear(2018);
}
return 0;
}