disarium.a60

b̲e̲g̲i̲n̲
    c̲o̲m̲m̲e̲n̲t̲ Find some Disarium numbers - numbers whose digit position-power sums
            are equal to the number, e.g. 135 = 1^1 + 3^2 + 5^3.
            Output: 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427;
    i̲n̲t̲e̲g̲e̲r̲ a̲r̲r̲a̲y̲ power [1 : 9, 0 : 9];
    i̲n̲t̲e̲g̲e̲r̲ count, powerOfTen, length, n, d;

    c̲o̲m̲m̲e̲n̲t̲ compute the nth powers of 0-9;
    f̲o̲r̲ d := 0 s̲t̲e̲p̲ 1 u̲n̲t̲i̲l̲ 9 d̲o̲
        power[1, d] := d;
    f̲o̲r̲ n := 2 s̲t̲e̲p̲ 1 u̲n̲t̲i̲l̲ 8 d̲o̲ b̲e̲g̲i̲n̲
        power[n, 0] := 0;
        f̲o̲r̲ d := 1 s̲t̲e̲p̲ 1 u̲n̲t̲i̲l̲ 9 d̲o̲
            power[n, d] := power[n - 1, d] * d
    e̲n̲d̲ n;

    c̲o̲m̲m̲e̲n̲t̲ print the first few Disarium numbers;
    count      :=  0;
    powerOfTen := 10;
    length     :=  1;
    n          := -1;
    f̲o̲r̲ n := n + 1 w̲h̲i̲l̲e̲ count < 18 d̲o̲ b̲e̲g̲i̲n̲
        i̲n̲t̲e̲g̲e̲r̲ v, dps, p;
        i̲f̲ n = powerOfTen t̲h̲e̲n̲ b̲e̲g̲i̲n̲
            c̲o̲m̲m̲e̲n̲t̲ the number of digfits just increased;
            powerOfTen := powerOfTen * 10;
            length     := length     +  1
        e̲n̲d̲;
        c̲o̲m̲m̲e̲n̲t̲ form the digit power sum;
        v   := n;
        dps := 0;
        f̲o̲r̲ p := length s̲t̲e̲p̲ -1 u̲n̲t̲i̲l̲ 1 d̲o̲ b̲e̲g̲i̲n̲
            dps := dps + power[p, v - (v ÷ 10 * 10)];
            v   := v ÷ 10
        e̲n̲d̲ p;
        i̲f̲ dps = n t̲h̲e̲n̲ b̲e̲g̲i̲n̲
            c̲o̲m̲m̲e̲n̲t̲ n is Disarium;
            count := count + 1;
            SPACE(1);
            print(n)
        e̲n̲d̲
    e̲n̲d̲ n;
    NLCR
e̲n̲d̲