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fortran66のブログ

fortran について書きます。

Problem 011

めんどくさいw 美しくない。

ソース・プログラム

この入力だと配列の列と行が反対になってる。

    program PEuler011
      implicit none
      integer, parameter :: n = 20
      integer, parameter :: itab(n, n) = [ & ! transpose
        [08,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,08],&
        [49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00],&
        [81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65],&
        [52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91],&
        [22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],&
        [24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50],&
        [32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],&
        [67,26,20,68,02,62,12,20,95,63,94,39,63,08,40,91,66,49,94,21],&
        [24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],&
        [21,36,23,09,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95],&
        [78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,09,53,56,92],&
        [16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57],&
        [86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58],&
        [19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40],&
        [04,52,08,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66],&
        [88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69],&
        [04,42,16,73,38,25,39,11,24,94,72,18,08,46,29,32,40,62,76,36],&
        [20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16],&
        [20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54],&
        [01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48] ]
      integer :: k, icol, irow, iprod(4)
      !holizontal
      iprod(1) = maxval([([(product(itab(irow, icol:icol + 3)), icol = 1, n - 3)], irow = 1, n)])
      !vertical
      iprod(2) = maxval([([(product(itab(irow:irow + 3, icol)), irow = 1, n - 3)], icol = 1, n)])
      !diagonal1 \
      iprod(3) = maxval([([(product([(itab(irow + k, icol + k), k = 0, 3)]), icol = 1, n - 3)], irow = 1, n - 3)])
      !diagonal2 /
      iprod(4) = maxval([([(product([(itab(irow + k, icol - k), k = 0, 3)]), icol = 4, n)], irow = 1, n - 3)])
      !
      print *, maxval(iprod) 
      stop
    end program PEuler011

実行結果

70600674
続行するには何かキーを押してください . . .