【codeforces 779B】Weird Rounding

【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;
}
posted @ 2017-10-04 18:45  AWCXV  阅读(376)  评论(0编辑  收藏  举报