1 条题解
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0
C :
#include<stdio.h> int main() { int Y,M,D,m[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}; int i,k,d; while(scanf("%d%d%d",&Y,&M,&D)!=EOF) { k=0; if((Y%4==0&&Y%100!=0)||(Y%400==0)) k=1; for(d=0,i=1;i<M;i++) d+=m[i][k]; d+=D; printf("%d\n",d); } return 0; }
C++ :
#include<stdio.h> int main() { int Y,M,D,m[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}; int i,k,d; while(scanf("%d%d%d",&Y,&M,&D)!=EOF) { k=0; if((Y%4==0&&Y%100!=0)||(Y%400==0)) k=1; for(d=0,i=1;i<M;i++) d+=m[i][k]; d+=D; printf("%d\n",d); } return 0; }
Pascal :
program t; const a:array[1..12]of integer=(31,28,31,30,31,30,31,31,30,31,30,31); b:array[1..12]of integer=(31,29,31,30,31,30,31,31,30,31,30,31); var y,m,d,i,s:longint; begin while not eof do begin read(y,m,d); s:=0; if (y mod 4=0)and(y mod 100<>0)or(y mod 400=0) then for i:=1 to m-1 do s:=s+b[i] else for i:=1 to m-1 do s:=s+a[i]; writeln(s+d); end; end.
- 1
信息
- ID
- 1837
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者