编程之美2.8 | 找符合条件的整数

思路还是相当地巧妙。

求余数的话，(a+b)%n=(a%n+b%n)%n;

用vector来表示整数的话（出现1的位置），可以避免溢出。

注意第20行，在更新remainders[(j+r)%n]时，要确保每个remainders的每个序列都是递增的，不能存在相等的情况。

#include <time.h>
#include <math.h>
#include <stdlib.h>

using namespace std;

long long findMInZeroOneNum(int n) {
        vector<vector<int> > remainders(n);
        remainders[1].push_back(0);

        bool update = false;
        int noupdate = 0;
        for (int i = 1, j = 10 % n; ; j = (j * 10) % n, i++) {
                if (remainders[j].empty()) {
                        remainders[j].push_back(i);
                        update = true;
                }
                for (int r = 1; r < n; ++r) {
                        if (remainders[r].empty()) continue;
                        if (i <= remainders[r].back()) continue;
                        if (!remainders[(j + r) % n].empty()) continue;
                        remainders[(j + r) % n] = remainders[r];
                        remainders[(j + r) % n].push_back(i);
                        update = true;
                }
                if (!update) noupdate++;
                else noupdate = 0;
                if (noupdate == n || !remainders[0].empty()) break;
        }

        if (remainders[0].empty()) return -1;
        else {
                long long ans = 0;
                for (int i = 0; i < remainders[0].size(); ++i) {
                        ans += pow(10, remainders[0][i]);
                }
                return ans;
        }
}

long long bruteForce(int n) {
        if (n <= 0) return -1;
        for (long long i = 1; ; ++i) {
                if (i % n != 0) continue;
                long long tmp = i;
                while (tmp) {
                        int d = tmp % 10; 
                        if (d != 0 && d != 1) break;
                        tmp /= 10;
                }
                if (tmp == 0) return i;
        }
        return -1;
}

int main() {
        srand(time(NULL));
        int n = rand() % 99;

        cout << n << endl;
        long long n1 = findMInZeroOneNum(n);
        cout << n1 << endl;
        long long n2 = bruteForce(n);
        cout << n2 << endl;
        if (n1 != n2) {
                cout << n << " " << n1 << " " << n2 << endl;
        }

        return 0;
}