1 solutions

  • 0
    @ 2024-12-10 21:47:36

    C :

    #include<stdio.h>
    int DayofYear(int year,int month,int day);
    int main()
    {
    	int y[100],m[100],d[100],num[100],l,i;
    	scanf("%d",&l);
    	for(i=0;i<l;i++)
    	{
    		scanf("%d-%d-%d",&y[i],&m[i],&d[i]);
    		num[i] = DayofYear(y[i],m[i],d[i]);
    	}
    	for(i=0;i<l;i++)
    	{
    		if(i!=l-1)
    		printf("%d\n",num[i]);
    		else
    		printf("%d",num[i]);
    	} 
    	
    	return 0;
    }
    
    int DayofYear(int x,int y,int z)
    {
    	int n;
    	if(x%4==0&&x%100!=0||x%400==0)
    	{
    		switch(y)
    		{
    			case 1:	n = z;break;
    			case 2:	n = 31+z;break;
    			case 3:	n = 60+z;break;
    			case 4:	n = 91+z;break;
    			case 5:	n = 121+z;break;
    			case 6:	n = 152+z;break;
    			case 7:	n = 182+z;break;
    			case 8:	n = 213+z;break;
    			case 9:	n = 244+z;break;
    			case 10:n = 274+z;break;
    			case 11:n = 305+z;break;
    			case 12:n = 335+z;break;
    		}
    	}
    	else
    	{
    		switch(y)
    		{
    			case 1:	n = z;break;
    			case 2:	n = 31+z;break;
    			case 3:	n = 59+z;break;
    			case 4:	n = 90+z;break;
    			case 5:	n = 120+z;break;
    			case 6:	n = 151+z;break;
    			case 7:	n = 181+z;break;
    			case 8:	n = 212+z;break;
    			case 9:	n = 243+z;break;
    			case 10:n = 273+z;break;
    			case 11:n = 304+z;break;
    			case 12:n = 334+z;break;
    		}
    	}
    	return n;
    }
    

    C++ :

    #include<stdio.h>
    int DayofYear(int year,int month,int day);
    int main()
    {
    	int y[100],m[100],d[100],num[100],l,i;
    	scanf("%d",&l);
    	for(i=0;i<l;i++)
    	{
    		scanf("%d-%d-%d",&y[i],&m[i],&d[i]);
    		num[i] = DayofYear(y[i],m[i],d[i]);
    	}
    	for(i=0;i<l;i++)
    	{
    		if(i!=l-1)
    		printf("%d\n",num[i]);
    		else
    		printf("%d",num[i]);
    	} 
    	
    	return 0;
    }
    
    int DayofYear(int x,int y,int z)
    {
    	int n;
    	if(x%4==0&&x%100!=0||x%400==0)
    	{
    		switch(y)
    		{
    			case 1:	n = z;break;
    			case 2:	n = 31+z;break;
    			case 3:	n = 60+z;break;
    			case 4:	n = 91+z;break;
    			case 5:	n = 121+z;break;
    			case 6:	n = 152+z;break;
    			case 7:	n = 182+z;break;
    			case 8:	n = 213+z;break;
    			case 9:	n = 244+z;break;
    			case 10:n = 274+z;break;
    			case 11:n = 305+z;break;
    			case 12:n = 335+z;break;
    		}
    	}
    	else
    	{
    		switch(y)
    		{
    			case 1:	n = z;break;
    			case 2:	n = 31+z;break;
    			case 3:	n = 59+z;break;
    			case 4:	n = 90+z;break;
    			case 5:	n = 120+z;break;
    			case 6:	n = 151+z;break;
    			case 7:	n = 181+z;break;
    			case 8:	n = 212+z;break;
    			case 9:	n = 243+z;break;
    			case 10:n = 273+z;break;
    			case 11:n = 304+z;break;
    			case 12:n = 334+z;break;
    		}
    	}
    	return n;
    }
    
    • 1

    【设计型】第9章: 指针 9.4(1)某月某日(1)

    Information

    ID
    502
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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