CodeForces 721B Passwords (水题)

Posted 2020-08-12 dwtfukgv

题意：给定 n 个密码，你要按长度不递减的顺序进行尝试，问你最多和最少试多少次可能找出密码，每尝试 k 次错误的，就要等5秒。

析：我们只要把长度全都统计下来，然后从1开始去找目标长度，最少的就是正好到目标长度，最多的就是把目标长度恰好试完。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[105];

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        string s;
        memset(a, 0, sizeof a);
        for(int i = 0; i < n; ++i){
            cin >> s;
            ++a[s.size()];
        }
        cin >> s;
        int cnt = s.size();
        int ans1 = 1, ans2 = 0;
        for(int i = 1; i < cnt; ++i)  ans1 += a[i];
        ans2 = ans1 + a[cnt] - 1;
        ans1 += ((ans1-1) / m) * 5;
        ans2 += ((ans2-1) / m) * 5;
        printf("%d %d\n", ans1, ans2);

    }
    return 0;
}