C language implementation, calculate the number of days of difference between two days

Given the year, month, and day of the two days, calculate the number of different days, and the year is small to large.

# Include <stdio. h> // for a given year, month, and day, calculate the number of days that have passed in the year int total_day (int year, int month, int day) {int sum = 0; switch (month) {case 1: sum = day; break; case 2: sum = 31 + day; break; case 3: sum = 59 + day; break; case 4: sum = 90 + day; break; case 5: sum = 120 + day; break; case 6: sum = 151 + day; break; case 7: sum = 181 + day; break; case 8: sum = 212 + day; break; case 9: sum = 243 + day; break; case 10: sum = 273 + d Ay; break; case 11: sum = 304 + day; break; case 12: sum = 334 + day; break; default: printf ("input month error \ n"); break;} if (month> 2) {if (year % 4 = 0) & (year % 100! = 0) | (year % 400) = 0) {sum = sum + 1 ;}} return sum ;} // number of days between the year and the year int total_year_day (int year1, int year2) {int sum_year_day = 0; int I = 0; sum_year_day = (year2-year1) * 365; for (I = year1; I <year2; I ++) {if (I % 4 = 0) & (I % 100! = 0) | (I % 400) = 0) {sum_year_day = sum_year_day + 1 ;}return sum_year_day;} int main () {int year1 = 2013, month1 = 1, day1 = 1; int year2 = 2013, month1 = 1, day2 = 1; int sum = 0; printf ("~~~~~~ Calculates the number of days for which the two days are different ~~~~~~ \ N "); printf (" Enter the start year, month, and day (Format: XXXX: XX) "); scanf (" % d: % d ", & year1, & month1, & day1); printf ("Enter the end year, month, and day (Format: XXXX: XX)"); scanf ("% d: % d ", & year2, & mon2, & day2); sum = total_year_day (year1, year2)-total_day (year1, month1, day1) + total_day (year2, mon2, day2); printf ("the number of days between them is % d \ n", sum); return 0 ;}