AtCoder Beginner Contest 068 （A B C D）

A - ABCxxx

 

水题。

#include <iostream>
using namespace std;
int main(){
    int n;
    cin>>n;
    cout << "ABC" << n << endl;
    return 0;
}

B - Break Number

水题。

#include <bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin >> n;
    int x = 1;
    while(x <= n) x<<=1;
    printf("%d\n",x>>1);
    return 0;
}

 

 

C - Cat Snuke and a Voyage

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

• 3≤N≤200 000
• 1≤M≤200 000
• 1≤a_i<b_i≤N
• (a_i,b_i)≠(1,N)
• If i≠j, (a_i,b_i)≠(a_j,b_j).

---

Input

Input is given from Standard Input in the following format:

N M
a1 b1
a2 b2
:
aM bM

Output

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

---

Sample Input 1

Copy

3 2
1 2
2 3

Sample Output 1

Copy

POSSIBLE

---

Sample Input 2

Copy

4 3
1 2
2 3
3 4

Sample Output 2

Copy

IMPOSSIBLE

You have to use three boat services to get to Island 4.

 

把与1和n相连的点分别放在两个集合里。

只要这两个集合就一个公共数就是POSSIBLE，否则就是IMPOSSIBLE。

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n, m;
    cin>>n>>m;
    set<int> st,st1;
    for(int i = 1; i <= m; i ++) {
        int x, y;
        cin >> x >> y;
        if(x == 1) st.insert(y);
        if(y == 1) st.insert(x);
        if(x == n) st1.insert(y);
        if(y == n) st1.insert(x);
    }
    set<int>::iterator it;
    for(it = st.begin(); it != st.end(); it ++) {
        if(st1.count((*it))){
            return 0*printf("POSSIBLE\n");
        }
    }
    printf("IMPOSSIBLE\n");
    return 0;
}

D - Decrease (Contestant ver.)

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N−1 or smaller.

• Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N−1 or smaller after a finite number of operations.

You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.

Constraints

• 0≤K≤50×10¹⁶

---

Input

Input is given from Standard Input in the following format:

K

Output

Print a solution in the following format:

N
a1 a2 ... aN

Here, 2≤N≤50 and 0≤a_i≤10¹⁶+1000 must hold.

---

Sample Input 1

Copy

0

Sample Output 1

Copy

4
3 3 3 3

---

Sample Input 2

Copy

1

Sample Output 2

Copy

3
1 0 3

---

Sample Input 3

Copy

2

Sample Output 3

Copy

2
2 2

The operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].

 

直接让N等于50，k=0是就是 0，1，2，4...49。k = 1时就是50，0，1，2,....48。k = 2 时就是49，50，0，1，2...47。50一个循环就都加1，k=50 就是1,2,3,...50

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll a[55];
int main() {
    ll n;
    cin >> n;
    for(int i = 1; i <= 50; i ++) {
        a[i] = n/50LL + i - 1;
    }
    n %= 50;
    for(int i = 1; 1LL*i <= n; i ++) {
        for(int j = 1; j <= 50; j ++)
            a[j]--;
        a[i] += 51;
    }
    cout << 50 << endl;
    for(int i = 1; i < 50; i ++)cout << a[i] << ' ';
    cout << a[50] << endl;
    return 0;
}