瓜瓜的A+B

#include <iostream> using namespace std;typedef long long ll;ll qmi(ll a, ll b, ll p) {ll res = 1;while (b){if (b & 1) res = res * a % p;a = a * a % p;b >>= 1;}return res; }int main() {int _; scanf("%d", &_);while (_ -- ){ll a, b, p; scanf("%lld%lld%lld", &a, &b, &p);printf("%lld\n", qmi((a + b) % p, p, p));} }

阿巴阿巴

#include <iostream> using namespace std;const int N = 1e3 + 10;int cnta[N], cntb[N];int main() {int _; scanf("%d", &_);while (_ -- ){int n;string s;cin >> n >> s;for (int i = 0; i < n; i ++ ){cnta[i] = cnta[i - 1] + (s[i] == 'a');cntb[i] = cntb[i - 1] + (s[i] == 'b');}bool ok = false;for (int l = 0; l + 7 < n; l ++ ){int r = l + 7;if (cnta[r] - cnta[l - 1] == 4 && cntb[r] - cntb[l - 1] == 4){ok = true;cout << l + 1 << ' ' << r + 1 << endl;break;}}if (!ok) cout << "impossible" << endl;for (int i = 0; i < n; i ++ ) cnta[i] = cntb[i] = 0;} }

俺拜俺拜

• 注意map如果ma[x]還不一定被賦值過的時候，一定要先count！然后在if里面再套一層if，不能&&
• 因為如果你用了ma[x]，直接自動給它賦值為0了，在此之后的count(x)操作全是true，因此，使用ma[x]需要謹慎（先用count）

#include <iostream> #include <unordered_map> using namespace std;int main() {int _; scanf("%d", &_);while (_ -- ){unordered_map<int, int> ma;int n, x; scanf("%d%d", &n, &x);string s; cin >> s;ma[0] = -1;bool ok = false;int cnt = 0;for (int i = 0; i < n; i ++ ){if (s[i] == 'a') cnt ++ ;else cnt -- ;if (ma.count(cnt)){if (i - (ma[cnt] + 1) + 1 >= x){ok = true;printf("%d %d\n", ma[cnt] + 2, i + 1);break;}}else ma[cnt] = i;}if (!ok) puts("impossible");} }

瓜瓜的特別任務

• 字符串哈希時注意 1.字符串下標要從1開始；2.p[0] = 1初始化

#include <iostream> using namespace std;typedef unsigned long long ull;const int N = 5e3 + 10, P = 131;ull h[N], p[N];ull get(int l, int r) {return h[r] - h[l - 1] * p[r - l + 1]; }int main() {int _; scanf("%d", &_);while (_ -- ){string s; cin >> s; s = " " + s;int n = (int)s.size();p[0] = 1;for (int i = 1; i <= n - 1; i ++ ){h[i] = h[i - 1] * P + s[i];p[i] = p[i - 1] * P;}int mx = 1;for (int len = 1; len <= n - 1; len ++ ){for (int i = 1; i + len - 1 <= n - 1; i ++ ){int cnt = 1, j = i + len;while (j + len - 1 <= n - 1 && get(i, i + len - 1) == get(j, j + len - 1)){cnt ++ ; j += len;}mx = max(mx, cnt);}}printf("%d\n", mx);} }

磊爺與傭兵戰記

#include <iostream> #include <cmath> #include <vector> using namespace std;typedef long long ll;const int N = 2e5 + 10;ll w[N]; ll sum[N]; bool is_lazy[N];vector<int> g[N]; vector<int> lazy[N];int main() {int n, m; scanf("%d%d", &n, &m);for (int i = 0, a, b; i < m && scanf("%d%d", &a, &b); i ++ ){g[a].push_back(b);g[b].push_back(a);}int t = sqrt(n);for (int i = 1; i <= n; i ++ ){if (g[i].size() >= t){is_lazy[i] = true;for (auto u : g[i])lazy[u].push_back(i);}}int q; scanf("%d", &q);int op, p;while (q -- ){scanf("%d%d", &op, &p);if (op == 1){int x; scanf("%d", &x);w[p] += x;if (is_lazy[p]) sum[p] += x;else{for (auto v : g[p])w[v] += x;}}else{ll res = w[p];for (auto v : lazy[p])res += sum[v];printf("%lld\n", res);}} }

瓜瓜的春天

• dp時注意是否要使用滾動數組
• 使用滾動數組的時候，注意，在初始化和最后輸出結果的時候也要&1

#include <iostream> using namespace std;typedef long long ll;const int N = 1e4 + 10; const ll mod = 998244353;ll f[2][N];ll qmi(ll a, ll b, ll p) {ll res = 1;while (b){if (b & 1) res = res * a % p;a = a * a % p;b >>= 1;}return res; }int main() {int ii, jj, a, b; scanf("%d%d%d%d", &ii, &jj, &a, &b);ll inv3 = qmi(3, mod - 2, mod);f[1 & 1][1] = inv3;for (int i = 1; i <= ii; i ++ )for (int j = 1; j <= jj; j ++ ){if (i == 1 && j == 1) continue;f[i & 1][j] = ((f[i & 1][j - 1] * inv3 % mod + f[i - 1 & 1][j] * inv3 % mod) % mod + f[i - 1 & 1][j - 1] * inv3 % mod) % mod;}printf("%lld\n", f[ii & 1][jj]); }

總結

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