引例

闰年：

bool isleapyear(int year)
{return (year%4==0&&year%400!==0)||(year%400==0);
}

例一：

bool isleapyear(int year)
{return (year%4==0&&year%400!=0)||(year%400==0);
}
int strtoint(char*str,int len)
{int i=0;int sum=0;for(i=0;i<len;i++){sum=sum*10+(str[i]-'0');    }return sum;
}           //将字符串型转换为整型数据；
int dayofyear(char*date)
{int sum=0;int monthday={31,28,31,30,31,30,31,31,30,31,30,31};    int year=0,month=0,day=0;year=strtoint(date,4);month=strtoint(date+5,2);//不要忘记-的长度；day=strtoint(date+8,2);monthday[1]=isleapyear(year)?29:28;for(i=1;i<month;i++){sum+=monthday[i-1]; }return sum+day;
}

题目一：

//思路是使用Kim larsen公式，也可以使用zelle公式
char * dayOfTheWeek(int day, int month, int year)
{if(month==1||month==2){month+=12;year--;}int w=(day+2*month+3*(month+1)/5+year+year/4-year/100+year/400+1)%7;char *arr[]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};return arr[w];
}

题目2：

方法简单粗暴

说好第二天发，凌晨三点也是第二天吧....................

最后这个题算同年的有bug，还请大佬们帮我看看；

跟着英雄学算法的第二天.................

int daysBetweenDates(char * date1, char * date2)
{int chartoint(char*date,int len){int sum=0;for(int i=0;i<len;i++){sum=sum*10+(date[i]-'0');}return sum;}bool isleapyear(int year){return (year%4==0&&year%400!=0)||(year%400==0);}int monthday[12]={31,28,31,30,31,30,31,31,30,31,30,31};int year1=chartoint(date1,4);int month1=chartoint(date1+5,2);int day1=chartoint(date1+8,2);int year2=chartoint(date2,4);int month2=chartoint(date2+5,2);int day2=chartoint(date2+8,1);int sum1=0;for(int i=1;i<month1;i++){monthday[1]=isleapyear(year1)?29:28;sum1+=monthday[i-1];}sum1=sum1+day1;//到本年1月1日的天数monthday[1]=28;int sum2=0;for(int j=1;j<month2;j++){monthday[1]=isleapyear(year2)?29:28;sum2+=monthday[j-1];}sum2=sum2+day2;//到本年1月1日的天数int yearday=0;if(year1>year2){for(int i=year2;i<year1;i++){int yearnum=isleapyear(i)?366:365;yearday+=yearnum;}yearday=yearday+sum1-sum2;}else if(year1==year2){if(sum1>sum2)yearday=sum1-sum2;elseyearday=sum2-sum1;}else{for(int i=year1;i<year2;i++){int yearnum=isleapyear(i)?366:365;yearday+=yearnum;}yearday=yearday+sum2-sum1;}return yearday;
}