AtCoder Beginner Contest 158

AtCoder Beginner Contest 158

https://atcoder.jp/contests/abc158
基础不牢，地动山摇

D - String Formation

一个小小的STL应用

#include <bits/stdc++.h>
#define ll long long

using namespace std;
string s;
int q, t, f;
char c;

int main () {
    cin >> s >> q;
    int tt = 0; //代表此刻状态
    stack <char> st;
    queue<char> ed;
    while (q --) {
        cin >> t;
        if (t == 2) {
            cin >> f >> c;
            if (f == 1) {
                if (tt == 1)    ed.push (c);
                else    st.push (c);
            }
            else {
                if (tt == 0)    ed.push (c);
                else    st.push (c);
            }
        }
        else    tt ^= 1;
    }
    string t;
    while (!st.empty ())    t += st.top (), st.pop ();
    t += s;
    while (!ed.empty ())    t += ed.front (), ed.pop ();    

    if (tt)    reverse (t.begin (), t.end ());
    cout << t;
}

E - Divisible Substring

是非常经典的数学小结论应用。
应该是利用了费马小定理，然后转化一下：

注意从右往左才是低位向高位

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 2e5 + 5;
int n, p, len;
string s;
ll ans, ss;

int main () {
    cin >> n >> p >> s;
    s = ' ' + s;
    if (p == 2 || p == 5) {
        for (int i = 1; i <= n; i++) {
            if ((s[i] - '0') % p == 0)  ans += i;
        }
        cout << ans << endl;
        return 0;
    }

    map<ll, int> mp;
    mp[0] = 1; 
    ll pw = 1;
    for (int i = n; i >= 1; i--) {
        ss = ss % p + (s[i] - '0') * pw % p;
        ss %= p, pw = pw * 10 % p;
        ans += mp[ss], mp[ss] ++;
    }
    cout << ans;
}

//费马小定理: a^{p-1}=1(mod p)
//10^{p-1}=1(mod p)
//(ai*10^{p-1}+aj) mod p = (ai+aj)*10^{p-1} mod p

//转化为有多少个区间和mod p = 0
//同余即可

F - Removing Robots

神奇的dp。
\(f_i\) 表示后 \(i\) 个的方案数，然后就从后往前更新，单调栈找第一个未被影响的数。

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 2e5 + 5, mod = 998244353;
pii p[N];
int n, f[N]; //后i个的方案数

int main () {
    cin >> n;
    for (int i = 1; i <= n; i++)    cin >> p[i].first >> p[i].second;
    sort (p + 1, p + n + 1);
    f[n+1] = 1;
    stack <pii> s;
    for (int i = n; i >= 1; i--) {
        int nxt = i + 1, t = p[i].first + p[i].second;
        while (!s.empty () && p[s.top().first].first < t) {
            nxt = s.top ().second; //nxt是不被影响的第一个
            s.pop ();
        }
        s.push ({i, nxt});
        f[i] = (f[i+1] + f[nxt]) % mod; 
    }
    cout << f[1];
}