結果
問題 | No.232 めぐるはめぐる (2) |
ユーザー | jj |
提出日時 | 2017-01-26 23:15:20 |
言語 | Fortran (gFortran 13.2.0) |
結果 |
AC
|
実行時間 | 62 ms / 1,000 ms |
コード長 | 2,193 bytes |
コンパイル時間 | 2,172 ms |
コンパイル使用メモリ | 29,464 KB |
実行使用メモリ | 4,372 KB |
最終ジャッジ日時 | 2023-10-12 13:47:49 |
合計ジャッジ時間 | 3,792 ms |
ジャッジサーバーID (参考情報) |
judge11 / judge12 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 25 ms
4,352 KB |
testcase_01 | AC | 62 ms
4,372 KB |
testcase_02 | AC | 25 ms
4,368 KB |
testcase_03 | AC | 37 ms
4,372 KB |
testcase_04 | AC | 1 ms
4,368 KB |
testcase_05 | AC | 1 ms
4,368 KB |
testcase_06 | AC | 1 ms
4,368 KB |
testcase_07 | AC | 1 ms
4,368 KB |
testcase_08 | AC | 40 ms
4,372 KB |
testcase_09 | AC | 1 ms
4,368 KB |
testcase_10 | AC | 1 ms
4,372 KB |
testcase_11 | AC | 1 ms
4,368 KB |
testcase_12 | AC | 1 ms
4,368 KB |
testcase_13 | AC | 1 ms
4,368 KB |
testcase_14 | AC | 1 ms
4,368 KB |
testcase_15 | AC | 1 ms
4,372 KB |
testcase_16 | AC | 1 ms
4,372 KB |
testcase_17 | AC | 1 ms
4,368 KB |
testcase_18 | AC | 1 ms
4,372 KB |
testcase_19 | AC | 1 ms
4,368 KB |
testcase_20 | AC | 1 ms
4,372 KB |
testcase_21 | AC | 1 ms
4,372 KB |
testcase_22 | AC | 1 ms
4,368 KB |
testcase_23 | AC | 1 ms
4,368 KB |
testcase_24 | AC | 1 ms
4,368 KB |
ソースコード
program main implicit none integer::T,A,B,i integer::maxab,up read *,T,A,B maxab = MAX(A,B) if(T.lt.maxab) then print '(a)', "NO" return else if(A.eq.0.and.B.eq.0) then if(T.eq.1) then print '(a)', "NO" else if(MOD(T,2).eq.0) then print '(a)', "YES" do i=1,T/2 print '(a)',">" print '(a)',"<" end do else print '(a)', "YES" do i=1,(T-3)/2 print '(a)',">" print '(a)',"<" end do print '(a)',">" print '(a)',"^" print '(a)',"<v" end if return else if(A.eq.1.and.B.eq.1) then print '(a)', "YES" if(MOD(T,2).eq.1) then do i=1,(T-1)/2 print '(a)',">" print '(a)',"<" end do print '(a)',"^>" else do i=1,(T-2)/2 print '(a)',">" print '(a)',"<" end do print '(a)',">" print '(a)',"^" end if else if(T.eq.maxab) then print '(a)', "YES" do i=1,maxab if(A.gt.0) write(*,'(a)',advance='no') "^" if(B.gt.0) write(*,'(a)',advance='no') ">" print *, A = A - 1 B = B - 1 end do return else if(MOD(T,2).eq.MOD(maxab,2)) then print '(a)', "YES" do i=1,maxab if(A.gt.0) write(*,'(a)',advance='no') "^" if(B.gt.0) write(*,'(a)',advance='no') ">" print *, A = A - 1 B = B - 1 end do do i=1,(T-maxab)/2 print '(a)', ">" print '(a)', "<" end do return else print '(a)', "YES" if(A.gt.B) then A = A - 1 up = 1 else B = B - 1 up = 0 endif maxab = MAX(A,B) do i=1,maxab if(A.gt.0) write(*,'(a)',advance='no') "^" if(B.gt.0) write(*,'(a)',advance='no') ">" print *, A = A - 1 B = B - 1 end do do i=1,(T-1-maxab)/2 print '(a)', ">" print '(a)', "<" end do if(up.eq.1) then print '(a)', ">" print '(a)', "<^" else print '(a)', "^" print '(a)', "v>" endif return end if end program main