【题目链接】:http://codeforces.com/contest/779/problem/B

【题意】



问你要删掉几个数字才能让原来的数字能够被10^k整除;

【题解】

/*

    数字的长度不大;

    让你删掉最小的数字个数

    使得这个数字能被10^k整除;

    搜索;

    枚举哪些数字被删掉了;

    bool记录;

    最后再全部乘起来;

    枚举删掉1个,删掉两个即可、3、4...;

    注意前导0只能有一个的情况就好

*/

【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%lld",&x)

typedef pair<int, int> pii;

typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };

const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };

const double pi = acos(-1.0);

const int N = 15;

string s;

int po_w[12],k,len,di,num;

bool bo[15];

void dfs(int x, int now)

{

    if (now >= num)

    {

        int nu = 0,t = 0,fir=-1;

        rep1(i, 1, len)

            if (bo[i])

            {

                if (fir == -1)

                    fir = s[i] - '0';

                t++;

                nu = nu * 10 + s[i] - '0';

            }

        if (fir == 0 && t > 1) return;

        if (nu%di == 0)

        {

            printf("%d\n", num);

            exit(0);

        }

        return;

    }

    if (x > len)

        return;

    bo[x] = false;

    dfs(x + 1, now + 1);

    bo[x] = true;

    dfs(x + 1, now);

}

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    po_w[0] = 1;

    rep1(i, 1, 9)

        po_w[i] = po_w[i - 1] * 10;

    cin >> s;

    len = s.size();

    s = ' ' + s;

    scanf("%d", &k);

    di = po_w[k];

    memset(bo, true, sizeof bo);

    for (num = 0;num <= len-1;num++)

        dfs(1, 0);

    return 0;

}