1 条题解
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0
C :
#include <stdio.h> #include <stdlib.h> int main() { int y,m,a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; scanf("%d%d",&y,&m); if(m!=2) printf("%d",a[m]); if(m==2) if(y%4==0 || y%400==0) printf("29"); else printf("28"); return 0; }
C++ :
#include <iostream> #include <cstdio> using namespace std; int main() { int year,month; cin>>year>>month; switch (month) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: cout<<31<<endl; break; case 4: case 6: case 9: case 11: cout<<30<<endl; break; default: //2月 if( (year%100 != 0 && year%4 == 0) || (year%400 == 0) ) { //如果是闰年,2月29天 cout<<29<<endl; } else { //如果是平年,2月28天 cout<<28<<endl; } break; } return 0; }
Pascal :
program xx; var year,month,days:integer; begin read(year,month); case month of 1,3,5,7,8,10,12: days:=31; 4,6,9,11 : days:=30; 2 : if (year mod 4=0)and(year mod 100 <>0) or(year mod 400=0) then days:=29 else days:=28 end; writeln(days); end.
- 1
信息
- ID
- 1057
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者