Codeforces #451 (Div. 2) A B C D E

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本文提供了 CodeForces 平台编号为 898 的五道题目的详细解答，包括 A. Rounding、B. Proper Nutrition、C. PhoneNumbers、D. AlarmClock 和 E. Squares and not squares。每道题都附有清晰的思路说明及 AC 代码示例。

A.Rounding

题目链接 http://codeforces.com/contest/898/problem/A

题解

手速下

AC代码

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

#define LL long long
#define CLR(a,b) memset(a,(b),sizeof(a))
#define INF 0x3f3f3f3f
const int MAXN = 1e3+10;

int main()
{
    int n;
    scanf("%d",&n);
    if(n == 5) printf("0\n");
    else if(n%10 >= 5) printf("%d\n",n+(10-n%10));
    else printf("%d\n",n-n%10);


return 0;
}

B.Proper Nutrition

题目链接 http://codeforces.com/contest/898/problem/B

题解

一元二次不定时求解比赛时直接抄了个扩展gcd 发现不用这么麻烦QAQ

AC代码

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

#define LL long long
int main()
{
    LL n, a, b;
    cin >> n >> a >> b;
    bool flag = false;
    for(int i = 0; i <= n; i++) {
        if((n-a*i)%b==0 &&((n-a*i)/b)>=0) {
            cout << "YES" << '\n';
            cout << i << " " << (n-a*i)/b << endl;
            flag = true; break;
        }
    }
    if(!flag) cout << "NO" << '\n';
return 0;
}

C.Phone Numbers

题目链接 http://codeforces.com/contest/898/problem/C

题解

非常恶心的一个字符串处理题目 vector 什么的随便搞搞 STL各种套

AC代码

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

int n, k, cnt = 0;
string s, ss;
vector<string> v, rss;
map<string,bool> mp,ne;
map<string,vector<string> > mm;
map<string,vector<string> > :: iterator it;

bool check(string s1, string s2)
{
    reverse(s1.begin(),s1.end());
    reverse(s2.begin(),s2.end());
    if(s1.size() >= s2.size()) return false;
    for(int i = 0; i < s1.size(); i++) {
        if(s1[i] != s2[i]) return false;
    }
    return true;
}

int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> s >> k;
        if(ne[s] == false) {
            cnt ++; ne[s] = true;
        }
        while(k--) {
            cin >> ss; mm[s].push_back(ss);
        }
    }
    cout << cnt << endl;
    for(it = mm.begin(); it != mm.end(); it++) {
        cout << it->first << ' ';
        v = it->second;
        rss.clear(); mp.clear();
        for(int i = 0; i < v.size(); i++) {
            if(mp[v[i]] == false) {
                bool flag = false;
                for(int j = 0; j < v.size(); j++) {
                    if(check(v[i],v[j])) {
                        flag = true; break;
                    }
                }
                if(flag == false) {
                    rss.push_back(v[i]);
                    mp[v[i]] = true;
                }
            }
        }
        cout << rss.size() << ' ';
        for(int i = 0; i < rss.size(); i++) {
            cout << rss[i] << ' ';
        }
        cout << "\n";
    }
return 0;
}

D. Alarm Clock

题目链接 http://codeforces.com/contest/898/problem/D

题解

一个简答的贪心不过写法要优美点不然有可能T，也可以从树状数组维护

AC代码

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int MAXN = 1e6+10;
int arr[MAXN];
int brr[MAXN];
int main()
{
    ios::sync_with_stdio(false);
    int n, m, k;
    cin >> n >> m >> k;
    for(int i = 0; i < n; i++) cin >> arr[i];
    sort(arr,arr+n);
    int h = 0,  l = 0;
    int ans = 0;
    for(int i = 0; i < n; i++) {
        brr[l++] = arr[i];
        while(brr[l-1]-brr[h]+1>m && l>h) h++;
        while(l-h+1>k) l--,ans++;
    }
    cout << ans <<endl;
return 0;
}

E. Squares and not squares

题目链接 http://codeforces.com/contest/898/problem/E

题解

维护两个vector就可以了可能是cf 最水的E了吧

AC代码

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

#define LL long long

int main()
{
    ios::sync_with_stdio(false);
    int n; vector<LL> v1, v2;
    cin >> n;
    int cnt = 0;
    for(int i = 0; i < n; i++) {
        LL x; cin >> x;
        LL y = sqrt(double(x));
        if(y*y == x) {
            cnt ++;
            if(x == 0) v1.push_back(2);
            else v1.push_back(1);
        }
        else {
            LL p = min(x-y*y,(y+1)*(y+1)-x);
            v2.push_back(p);
        }
    }
    n = n/2;
    if(cnt == n) cout << 0 << endl;
    else if(cnt > n) {
        sort(v1.begin(),v1.end());
        LL ans = 0, i =0;
        while(cnt > n) ans+=v1[i++], cnt--;
        cout << ans << endl;
    }
    else {
        sort(v2.begin(),v2.end());
        LL ans = 0, i = 0;
        while(cnt < n) ans+=v2[i++],cnt++;
        cout << ans << endl;
    }

return 0;
}