【解決方法】このコードのどこが間違っているのでしょうか?

C プログラムを使用して年齢を日数で取得します。

私が試したこと：

#include<stdio.h>
int main(){
	int date,month,year,cur_year,cur_date,cur_month,i,n,s=0,k,p,m,a[7]={1,3,5,7,8,10,12},b[4]={4,6,9,11};
	printf("Enter date of birth: ");
	scanf("%d",&date);
	printf("Enter month of birth in number: ");
	scanf("%d",&month);
	printf("Enter year of birth: ");
	scanf("%d",&year);
	printf("Enter current date: ");
	scanf("%d",&cur_date);
	printf("Enter current month: ");
	scanf("%d",&cur_month);
	printf("Enter current year: ");
	scanf("%d",&cur_year);
               
           //for getting number of days remained in your birth month

	for(i=0;i<=10;i++){
		if(month==a[i])
		p = 31 - date;
		if(month==b[i])
		p = 30 - date;
	}
	if(month==2){
		if(year%4==0)
		p = 29 - date;
		else
		p = 28 - date;
	}

              //for getting number of days in remaining months in the birth year

	switch(month){
		case 1:{
			s = (31*6)+(30*4);
			if(year%4==0)
			k = s + 29;
			else
			k = s + 28;
			break;
		}
		case 2:{
			k = (31*6)+(30*4);
			break;
		}
		case 3:{
			k = (31*5)+(30*4);
			break;
		}
		case 4:{
			k = (31*5)+(30*4);
			break;
		}
		case 5:{
			k = (31*4)+(30*3);
			break;
		}
		case 6:{
			k = (31*4)+(30*2);
			break;
		}
		case 7:{
			k = (31*3)+(30*2);
			break;
		}
		case 8:{
			k = (31*2)+(30*2);
			break;
		}
		case 9:{
			k = (31*2)+30;
			break;
		}
		case 10:{
			k = 30 + 31;
			break;
		}
		case 11:{
			k = 31;
			break;
		}
	}


                //for getting number of days in years after your birth year and before present year

	for(n=year+1;n<=cur_year-1;n++){
		if(n%4==0)
		s=s+366;
		else
		s=s+365;
	}

              //for getting number of days passed in the current year till todays month

	switch(cur_month){
		case 12:{
			if(cur_year%4==0)
			m = (31*6)+(30*4)+29;
			else
			m = (31*6)+(30*4)+28;
			break;
		}
		case 11:{
			if(cur_year%4==0)
			m = (31*6)+(30*3)+29;
			else
			m = (31*6)+(30*3)+28;
			break;
		}
		case 10:{
			if(cur_year%4==0)
			m = (31*5)+(30*3)+29;
			else
			m = (31*5)+(30*3)+28;
			break;
		}
		case 9:{
			if(cur_year%4==0)
			m = (31*5)+(30*2)+29;
			else
			m = (31*5)+(30*2)+28;
			break;
		}
		case 8:{
			if(cur_year%4==0)
			m = (31*4)+(30*2)+29;
			else
			m = (31*4)+(30*2)+28;
			break;
		}
		case 7:{
			if(cur_year%4==0)
			m = (31*3)+(30*2)+29;
			else
			m = (31*3)+(30*2)+28;
			break;
		}
		case 6:{
			if(cur_year%4==0)
			m = (31*3)+(30*1)+29;
			else
			m = (31*3)+(30*1)+28;
			break;
		}
		case 5:{
			if(cur_year%4==0)
			m = (31*2)+(30*1)+29;
			else
			m = (31*2)+(30*1)+28;
			break;
		}
		case 4:{
			if(cur_year%4==0)
			m = (31*2)+29;
			else
			m = (31*2)+28;
			break;
		}
		case 3:{
			if(cur_year%4==0)
			m = (31*1)+29;
			else
			m = (31*1)+28;
			break;
		}
		case 2:{
			m = 31;
			break;
		}
	}

         //number of months passed in the currents month is included below in the sum

	printf("Current age in days is %d",p+k+s+m+cur_date);
	return 0;
}