A Sanitize Hands 

问题：

思路：前缀和，暴力，你想咋做就咋做

代码：

#include <iostream>using namespace std;const int N = 2e5 + 10;int n, m;
int a[N];int main() {cin >> n >> m;for(int i = 1; i <= n; i ++ ) {cin >> a[i];}int ans = 0;for(int i = 1; i <= n; i ++ ) {m -= a[i];ans = i;if(m <= 0) break;}if(m < 0) cout << ans - 1;else cout << ans;return 0;    
}

B Uppercase and Lowercase

问题：

思路：大小写转换，这里有个问题，为什么我的转换最后都变成数字了，先留个疑问

代码：

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 2e5 + 10;string str;int main() {cin >> str;int cnt1 = 0, cnt2 = 0;for(auto t: str) {if(t >= 'a' && t <= 'z') cnt1 ++;else cnt2 ++;}if(cnt1 >= cnt2)transform(str.begin(),str.end(),str.begin(),::tolower);else transform(str.begin(),str.end(),str.begin(),::toupper);cout<<str<<endl;return 0;
}

C Sierpinski carpet

问题：

思路：阴间题，第一眼递归，但是不想求太多坐标，于是想到把图全变成‘#’最后填充'.'

代码：

#include <iostream>
#include <cmath>
#include <vector>using namespace std;const int N = pow(3, 6) + 10;char g[N][N];
int n;int main() {cin >> n;int len = pow(3, n);for(int i = 1; i <= len; i ++ ) {for(int j = 1; j <= len; j ++ ) {g[i][j] = '#';}}for(int level = 1; level <= n; level ++ ) {for(int i = 1 + pow(3, level - 1); i <= len; i += pow(3, level)) {for(int j = 1 + pow(3, level - 1); j <= len; j += pow(3, level)) {for(int k = i; k <= i + pow(3, level - 1) - 1; k ++ ) {for(int u = j; u <= j + pow(3, level - 1) - 1; u ++ ) {g[k][u] = '.';}}}}}for(int i = 1; i <= len; i ++ ) {for(int j = 1; j <= len; j ++ ) {cout << g[i][j];}cout << endl;}return 0;
}

D 88888888

问题：

思路：逆元，快速幂，对原式子变形后发现最后的结果实际上就是x 乘上一个等比数列，这是碰见的第一道逆元的题目，也明确了我对逆元的认识，由于 a / b % mod != (a % mod/ b % mod) % mod，而直接除的话会造成精度丢失，因此我们可以把除法变成乘法，根据费马小定理如果b和p互质，那么b的逆元就等于b ^ p - 2 因此可以快速幂求逆元

代码：
 

#include <iostream>using namespace std;const int mod = 998244353;long long x;int get(long long a) {int cnt = 0;while(a) {a /= 10;cnt ++;}return cnt;
}long long qmi(long long a, long long b) {long long res = 1;while(b) {if(b & 1) res = ((res % mod) * (a % mod)) % mod;b >>= 1;a = (a % mod * a % mod) % mod;}return res;
}int main() {cin >> x;int len = get(x);long long part1 = x % mod;long long a = qmi(10, (long long)len);long long b = qmi(a, x);b --;long long c = qmi(a - 1, 998244353 - 2);long long part2 = (b % mod * c % mod) % mod;cout << (part1 * part2) % mod;return 0;
}

E Reachability in Functional Graph

问题：

思路：考虑如果题目是一颗树的话那么直接一个记忆化即可，但是该题会出现环，因此考虑缩点，记得开long long

据说这是基环树板子，回头学一下基环树

代码：

#include <iostream>
#include <cstring>
#include <stack>
#include <map>using namespace std;const int N = (2e5 + 10) * 2;stack<int> stk;
int n;
int val[N], ne[N], h[N], idx;
int dfn[N], low[N], id[N], _size[N], scc_cnt, ts;
int cnt[N];
bool ins[N], st[N];
long long ans = 0;void add(int a, int b) {val[idx] = b;ne[idx] = h[a];h[a] = idx ++;
}void tarjan(int u) {dfn[u] = low[u] = ++ ts;stk.push(u);ins[u] = true;for(int i = h[u]; i != -1; i = ne[i]) {int j = val[i];if(!dfn[j]) {tarjan(j);low[u] = min(low[u], low[j]);} else if(ins[j]) low[u] = min(low[u], dfn[j]);}if(dfn[u] == low[u]) {++ scc_cnt;int y;do {y = stk.top();stk.pop();ins[y] = false;id[y] = scc_cnt;_size[scc_cnt] ++;} while (y != u);}
}void dfs(int u) {for(int i = h[u]; i != -1; i = ne[i]) {int j = val[i];if(!st[j]) {dfs(j);st[j] = true;}cnt[u] += cnt[j];ans += _size[u] * cnt[j];}
}int main() {memset(h, -1, sizeof h);cin >> n;scc_cnt = n;for(int i = 1; i <= n; i ++ ) {int x;cin >> x;add(i, x);}for(int i = 1; i <= n; i ++ ) if(!dfn[i]) tarjan(i);for(int i = 1; i <= n; i ++ ) cnt[id[i]] = _size[id[i]];map<pair<int, int>, int> ma;for(int i = 1; i <= n; i ++ ) {for(int j = h[i]; j != -1; j = ne[j]) {int k = val[j];if(id[i] != id[k] && !ma[{i, k}]) {add(id[i], id[k]);ma[{i, k}] ++;}}}memset(st, 0, sizeof st);for(int i = scc_cnt; i > n; i -- ) {if(!st[i]) {st[i] = true;dfs(i);}}for(int i = scc_cnt; i > n; i -- ) ans += (long long)_size[i] * (_size[i] - 1);cout << ans + n;return 0;
}

F