Codeforces 617A Elephant 【水题】

最小步骤到达目标距离

最新推荐文章于 2021-10-07 10:58:00 发布

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本文讨论了一种算法，用于计算大象以多种步长移动时，从起点0到达目标位置x所需的最少步骤数。

An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at pointx(x > 0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.

Input

The first line of the input contains an integer x (1 ≤ x ≤ 1 000 000) — The coordinate of the friend's house.

Output

Print the minimum number of steps that elephant needs to make to get from point 0 to point x.

Sample test(s)

input
5

output
1

input
12

output
3

Note

In the first sample the elephant needs to make one step of length 5 to reach the point x.

In the second sample the elephant can get to point x if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach x in less than three moves.

题意：每次你可以走1、2、3、4、5步，问你从0到x最少需要多少步。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (1000+10)
#define MAXM (200000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
using namespace std;
int main()
{
    int x; Ri(x);
    int num = x / 5 + 1;
    if(x % 5 == 0)
        num--;
    Pi(num);
    return 0;
}