AtCoder Beginner Contest 119

目录

• A - Still TBD
• B - Digital Gifts
• C - Synthetic Kadomatsu
• D - Lazy Faith

A - Still TBD

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;

int main(){
    string s;
    cin >> s;
    if (s[5] == '1') cout << "TBD" << endl;
    else if(s[6]>'4')
        cout << "TBD" << endl;
    else
        cout << "Heisei" << endl;
    return 0;
}

B - Digital Gifts

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n;
double res = 0;
int main(){
    cin >> n;
    for (int i = 0; i < n;i++){
        double x;
        string s;
        cin >> x >> s;
        if (s == "JPY") res += x;
        else
            res += 380000.0 * x;
    }
    cout << res << endl;
    return 0;
}

C - Synthetic Kadomatsu

给出abc三个数，以及n个备选的数，每次可以花费1点魔法值，讲一个数+1，也可以花费10点魔法值，将两个数相加，问最少需要多少魔法值，获得最终的abc三个数，n<=8

dfs暴力算即可

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, a, b, c;
int l[N];
int res = 0x3f3f3f3f;
void dfs(int now, int nowa, int nowb, int nowc, int add) {
    if (now == n) {
        if (nowa == a || nowb == b || nowc == c) return;
        res = min(res, abs(nowa) + abs(nowb) + abs(nowc) + add);
        return;
    }
    dfs(now + 1, nowa - l[now], nowb, nowc, add + (nowa == a ? 0 : 10));
    dfs(now + 1, nowa, nowb - l[now], nowc, add + (nowb == b ? 0 : 10));
    dfs(now + 1, nowa, nowb, nowc - l[now], add + (nowc == c ? 0 : 10));
    dfs(now + 1, nowa, nowb, nowc, add);
}

int main() {
    cin >> n >> a >> b >> c;
    for (int i = 0; i < n; i++) cin >> l[i];
    dfs(0, a, b, c, 0);
    cout << res << endl;
    return 0;
}

D - Lazy Faith

在数轴上有a和b两类点，现在给出一个点k，问从k点开始，走到最近的a点和b点一共需要走多远

对于输入的k，先lower_bound找到大于等于k的a点和b点，枚举走这两个点或者走较小的点的情况即可，一共四种情况

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
LL a[N], b[N];

int q, n, m;
int main() {
    cin >> n >> m >> q;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < m; i++) cin >> b[i];
    while (q--) {
        LL now;
        cin >> now;
        int posa = lower_bound(a, a + n, now) - a;
        int posb = lower_bound(b, b + m, now) - b;
        LL res = 0x3f3f3f3f3f3f3f3f;
        if (posa > 0 && posb < m) {
            res = min(res, b[posb] - a[posa - 1] +
                               min(now - a[posa - 1], b[posb] - now));
        }
        if (posa < n && posb > 0) {
            res = min(res, a[posa] - b[posb - 1] +
                               min(now - b[posb - 1], a[posa] - now));
        }
        if(posa<n&&posb<m){
            res = min(res, max(a[posa], b[posb]) - now);
        }
        if(posa>0&&posb>0){
            res = min(res, now - min(a[posa - 1], b[posb - 1]));
        }
        cout << res << endl;
    }
    return 0;
}