itsa68 4算數移位相加

移位 例:000123 => 123000 可看成 (向左移*10 向右移*0.1)
原 1*10^2+2*10^1+3*10^0
果 1*10^5+2*10^4+3*10^3 = (1*10^2+2*10^1+3*10^0)*10^3
現在有一個數n，4位數，向右移3次後相加，可寫成
S = n+n*0.1+n*0.01+n*0.001
= n*(1+0.1+0.01+0.001)
= n*1.111
n = S / 1.111
算出來會是除了個位數的其他位數
因為有誤差所以要加回去(進位等)
跑0~20，加完後，判斷移位相加後的結果與S是否相同，即為所求a0

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#include<iostream> #include<string> #include<cmath> #include<iomanip> using namespace std; bool check(unsigned long long n, unsigned long long s, unsigned long long len) { unsigned long long sum = 0; for (unsigned long long i = 0; i < len; i++) { sum += n; n /= 10; } if (sum - s == 0) return true; return false; } int main() { int found; unsigned long long n, len, res; double div, fr; while (cin >> n) { len = (unsigned long long) (log10(n)) + 1; div = 1; fr = 0.1; found = 0; for (int i = 2; i <= len; i++) { div += fr; fr *= 0.1; } res = (unsigned long long)(n / div); for (int j = 0; j <= 20; j++) { if (check(res + j, n, len)) { cout << res + j << endl; found = 1; break; } } if (found == 0) cout << -1 << endl; } return 0; }