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Perfect number

C++ Loops

sum of divisors

Flowchart (ISO 5807)

YesNoYesNoStartInput nsum = 0i = 1, n - 1, 1n % i == 0sum += isum == nOutput «Досконале»Output «Ні»EndFigure 1 — perfect

Source code

void perfect(int n) {
    int sum = 0;
    for (int i = 1; i < n; i++) {
        if (n % i == 0) {
            sum += i;
        }
    }
    if (sum == n) {
        cout << "Досконале";
    } else {
        cout << "Ні";
    }
}