31+29+31+30+31+30=
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31+29+31+30+31+30=
31+29+31+30+31+30=
31+29+31+30+31+30=
结果等于182
31+29+31+30+31+30=
第31,30,29,
29,30,31
29 30 31
main(){int a[]={0,31,28,31,30,31,30,31,31,30,31,30,31};/*平年的每月最后一天*/int b[]={0,31,29,31,30,31,30,31,31,30,31,30,31};/*闰年的每月最后一天*/int year,month,date;int year1,month1,date1;int year2,month2,date2;int d,m,day,i=1;lon
关于C语言编写一万年历的问题,只能拿出50财富聊表谢意了#include#includeint main(void){intyear,month,day;intdays=0,i,j,n,m,k,o;intm1[]={31,28,31,30,31,30,31,31,30,31,30,31};intm2[]={31,29,31,30,31,30,31,31,30,31,30,31};time_t toda
求大神将下面的c语言改成能在Matlab上运行的!急!急!#include intdays[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31}, {0,31,29,31,30,31,30,31,31,30,31,30,31}};char s[8][10] ={Monday,Tuesday,Wednesday,Thursday,F
[(1+29)*(1+29/2)*...*(1+29/30)*(1+29/31)]/[(1+31)*(1+31/2)*(1+31/3)*...*(1+31/28)*(1+31/29)=?[(1+29)*(1+29/2)*...*(1+29/30)*(1+29/31)]/[(1+31)*(1+31/2)*(1+31/3)*...*(1+31/28)*(1+31/29)〕=?上面那道题少了一个中括号
28,29 30 31 32
29*31*(30^2+1)
27、28、29、30、31、
11+30+31 +30 +31+31+30+31 +30 +31+31 +28 +31+30 +31+30 +31 +13 等于多少?
30,31
30,31,
30,31,
int leapdays[13]={0,31,29,31,30,31,30,31,31,30,31,30,31}; int a[13]; int sum,total; sum=0; a[0]=0;sint leapdays[13]={0,31,29,31,30,31,30,31,31,30,31,30,31}; int a[13]; int sum,total; sum=0; a[0]=0;sum=0; for(int i=1;i
(24/23+30/29+32/31)/(12/23+15/29+16/31) =
代码24行出现 non-lvalue in assignment#include#define IS(x)x%100!=0&&x%4=0||x%400=0?1:0int dayofmonth[13][2]={0,0,31,31,28,29,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31,};struct Date {int year;int month;int day;void nextday(){day