1 条题解

  • 0
    @ 2025-4-12 21:33:37

    Pascal :

    var k,i : integer;
        d,d1 : real;
        dis,oil : array[0..100] of real;
    begin
      k := 1;  d := 500;                  { 从i=1处开始向始点倒推}
      dis[1] := 500;
      oil[1] := 500;
      repeat
        k := k+1;  d := d+500/(2*k-1);
        dis[k] := d;
        oil[k] := oil[k-1]+500;
      until d>=1000;
      dis[k] := 1000;                    {置始点至终点的距离值}
      d1 := 1000-dis[k-1];               {求贮油点k处至始点的距离}
      oil[k] := d1*(2*k+1)+oil[k-1];     {求始点藏油量}
      writeln('  No.        Distance                Oil');
      for i := 0 to k do     {输出第i个贮油点的距离为1000-dis[k-i],藏油量为oil[k-i]}
        writeln(i : 4,1000-dis[k-i] : 20: 10,oil[k-i] : 20 : 10);
    end.
    
    • 1

    信息

    ID
    261
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
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    递交数
    0
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