abc370 A-E題解 (AtCoder Beginner Contest 370)

A

這類簡單題，看清楚：

Output

Print Yes, No, or Invalid according to the instructions in the problem statement.

B

C

std，這樣寫（0->n-1，n-1->0），可以少寫一個vector

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

int main() {
    string s, t;
    cin >> s >> t;
    vector<string> ans;
    vector<int> v;
    int n = s.size();
    for (int i = 0; i < n; i++) {
        if (s[i] > t[i]) v.push_back(i);
    }
    for (int i = n - 1; i >= 0; i--) {
        if (s[i] < t[i]) v.push_back(i);
    }
    int sz = v.size();
    for (int i = 0; i < sz; i++) {
        s[v[i]] = t[v[i]];
        ans.push_back(s);
    }
    cout << sz << '\n';
    for (string e : ans) cout << e << '\n';
}

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long

const LL mod_1=1e9+7;
const LL mod_2=998244353;

const double eps_1=1e-5;
const double eps_2=1e-10;

const int maxn=2e5+10;

vector<int> v1,v2;

int main()
{
    int i,cnt=0;
    string s1,s2;
    cin>>s1>>s2;
    for (i=0;i<s1.size();i++)
        if (s1[i]>s2[i])
            v1.emplace_back(i), cnt++;
        else if (s1[i]<s2[i])
            v2.emplace_back(i), cnt++;
    cout<<cnt<<endl;
    for (auto d:v1)
    {
        s1[d] = s2[d];
        cout<<s1<<endl;
    }
    reverse(v2.begin(),v2.end());
    for (auto d:v2)
    {
        s1[d] = s2[d];
        cout<<s1<<endl;
    }

    return 0;
}

D

E

就是要找出前面

如果暴力做，

特殊情況，全部0 0 0 ； -1 1 -1 1 這樣的

好短程式碼，很快能寫完。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long

const LL mod=998244353;

const double eps_1=1e-5;
const double eps_2=1e-10;

const int maxn=2e5+10;

map<LL, LL> mp;

int main()
{
    LL n,k,a,i,ans=0,sum=0, exc_ans, exc_sum;
    mp[0]=1;
    sum=1;
    cin>>n>>k;
    for (i=1;i<=n;i++)
    {
        cin>>a;
        ans+=a;

        exc_ans = ans-k;

        if (mp.find(exc_ans)!=mp.end())
            exc_sum = (sum - mp[exc_ans] + mod) % mod;
        else
            exc_sum = sum;

        mp[ans] = (mp[ans] + exc_sum) % mod;
        sum = (sum + exc_sum) % mod;

        if (i==n)
            cout << (exc_sum % mod + mod) % mod;
    }

    return 0;
}

F

G