暂时咕着，慢慢补...赛时几题先放上来

A

打个表即可

#include<bits/stdc++.h>
using namespace std;
int main() {long long n;scanf("%lld",&n);if(n & 1) puts("1"); else puts("-1");
}

B

枚举端点，hash判断即可。

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
#define N 10100
#define base 13131char s[N];
ull h1[N], p[N], h2[N];
int n, ans = 0;ull gh1(int l, int r) { return h1[r] - h1[l - 1] * p[r - l + 1]; }
ull gh2(int l, int r) { return h2[l] - h2[r + 1] * p[r - l + 1]; }int main() {scanf("%s", s + 1); p[0] = 1;n = strlen(s + 1);for(int i = 1; i <= n; ++i) s[i + n] = s[i];n *= 2;for(int i = 1; i <= n; ++i) {h1[i] = h1[i - 1] * base + s[i];p[i] = p[i - 1] * base;}for(int i = n; i; i--) h2[i] = h2[i + 1] * base + s[i];for(int i = 1; i <= n; ++i) {for(int j = i; j - i + 1 <= n / 2 && j <= n; ++j) {int mid = (i + j) >> 1;if((j - i + 1) & 1 && gh1(i, mid - 1) == gh2(mid + 1, j)) ans = max(ans, j - i + 1);if(!((j - i + 1) & 1) && gh1(i, mid) == gh2(mid + 1, j)) ans = max(ans, j - i + 1);}}printf("%d\n", ans);
}

C

考虑答案长什么样子
\[ \frac{钱数大于3*n的方案数}{总方案数} \]
这个方案数dp一下就可以了。类似背包。
然后总方案数是什么呢？考虑每次一共有4种选择，相乘即\(4^n\)。
用个gcd约分即可。

#include <bits/stdc++.h>
using namespace std;
#define ll long longll f[40][130];
int n;int main() {cin >> n;f[0][0] = 1;for(int i = 1; i <= n; ++i) {for(int j = 1; j <= 4 * n; ++j) for(int v = 1; v <= 4; ++v) if(j - v >= 0) f[i][j] += f[i - 1][j - v];}ll m1 = 0;for(int i = 3 * n; i <= 4 * n; ++i) m1 += f[n][i];ll m2 = pow(4, n), g = __gcd(m1, m2);printf("%lld/%lld\n", m1 / g, m2 / g);
}

D

维护前后缀和，枚举不选点

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 5000010ll a[N], l[N], r[N];int main() {int n; cin >> n;for(int i = 1; i <= n; ++i) scanf("%lld", &a[i]), l[i] = l[i - 1] | a[i];for(int i = n; i; --i) r[i] = r[i + 1] | a[i];ll ans = 0;for(int i = 1; i <= n; ++i) {ans = max(ans, l[i - 1] | r[i + 1]);}  printf("%lld\n", ans);
}

E

倍增floyd，转移的时候相乘即可。注意细节。

#include <cstdio>
#include <algorithm>
#include <cstring>
#define ll long long
#define inf 0x3f3f3f3f
#define il inlinenamespace io {#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')#define I_int ll
inline I_int read() {I_int x = 0, f = 1;char c = getchar();while (c < '0' || c > '9') {if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return x * f;
}
char F[200];
inline void write(I_int x) {if (x == 0) return (void) (putchar('0'));I_int tmp = x > 0 ? x : -x;if (x < 0) putchar('-');int cnt = 0;while (tmp > 0) {F[cnt++] = tmp % 10 + '0';tmp /= 10;}while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int}
using namespace io;using namespace std;#define N 110
const ll mod = 1000000007;ll K,m,s,n;struct mat {int m[N][N];mat operator * (const mat x) const {mat c; memset(c.m, 0, sizeof(c.m));for(int k = 1; k <= n; ++k) {for(int i = 1; i <= n; ++i) {for(int j = 1; j <= n; ++j) {c.m[i][j] += m[i][k] * x.m[k][j] % mod;c.m[i][j] %= mod;}}}return c;}
}d[32];int main() {in(n);in(m);in(K);in(s); int tot = 0;for(int i = 1; i <= m; ++i) {int x, y; in(x), in(y);d[0].m[x][y]++;}for(int i = 1; (1 << i) <= K; ++i) d[i] = d[i - 1] * d[i - 1];mat ans;--K; ans = d[0];for(int i = 0; (1 << i) <= K; ++i) if((K>>i)&1) ans = ans * d[i];ll sum = 0;for(int i = 1; i <= n; ++i) {if(i == s) continue;if(ans.m[s][i]!=-1) sum += ans.m[s][i] % mod;sum %= mod;}outn(sum);
}

H

单调栈板子。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 1000100
int a[N], n, k, h[N], w[N], s[N];
ll ans = 0;int main() {scanf("%d", &n);for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);for(int i = 1; i <= n; ++i) scanf("%d", &h[i]), ans = max(ans, (ll)h[i] * a[i]);int top = 0;for(int i = 1; i <= n; ++i) {if(a[i] > h[s[top]]) {s[++top] = i, w[top] = 1;} else {ll W = 0;while(top && h[s[top]] > h[i]) {W += w[top];ans = max(ans, (ll)W * h[s[top]]);--top;}s[++top] = i; w[top] = W + a[i];}}printf("%lld\n", ans);
}

I

做一下nim，如果必败就枚举要取哪个，如果都不行那就还是必败

#include <bits/stdc++.h>
using namespace std;
#define N 100010
int a[N], s, n, k;int main() {cin >> n >> k;for(int i = 1; i <= n; ++i) {cin >> a[i]; s ^= a[i];}if(s) return puts("YES"), 0;for(int i = 1; i <= n; ++i) {int now = s ^ a[i];if(a[i] >= k && now ^ (a[i] - k)) return puts("YES"), 0;}puts("NO");
}

J

套路莫反。这里放上推导。复杂度\(O(\sqrt{n})\)
\[ \begin{aligned} &\sum \sum (i,j)^2\\ &=\sum^n d^2\sum \sum[(i,j)=d]\\ &=\sum^n d^2 \sum_{i=1}^{n/d}\sum_{j=1}^{m/d}[(i,j)=1]\\ &=\sum^n d^2 \sum_{k=1}^{n/d}\mu(k)\frac{n}{kd}\frac{m}{kd}\\ &T = kd\\ &=\sum_{T=1}^n\frac{n}{T}\frac{m}{T}\sum_{k|T}\mu(k)(\frac{T}{k})^2\\ &id^2*\mu\\ &=F(1)=1, F(p)=p^2-1, F(p^2)=F(p)*p^2 \end{aligned} \]

\[ \begin{aligned} &\sum_{k|T}\mu(k)(\frac{T}{k})^2\\ &\mu*id^2\\ &F(1)=1\\ &F(p)=1*(p/1)^2+(-1)*(p/p)^2=p^2-1\\ &F(p^2)=1*(p^2/1)^2+(-1)*p^2+0=(p^2)^2-p^2=(p^2-1)*p^2 \end{aligned} \]

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 1000100
#define M 1000000
const ll mod = 1000000007;ll n, m;
int p[N], cnt, vis[N];
ll F[N];void init() {F[1] = 1;for(int i = 2; i <= M; ++i) {if(!vis[i]) p[++cnt] = i, F[i] = ((ll)i * i - 1ll) % mod;for(int j = 1; i * p[j] <= M && j <= cnt; ++j) {vis[i * p[j]] = 1;if(i % p[j] == 0) {F[i * p[j]] = F[i] * (p[j] * p[j] % mod) % mod;break;}F[i * p[j]] = (ll)F[i] * F[p[j]] % mod;}}for(int i = 1; i <= M; ++i) F[i] += F[i - 1], F[i] %= mod;
}int main() {init();scanf("%lld%lld", &n, &m);if(n > m) swap(n, m);ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans = (ans + (ll)(n / l) * (m / l) % mod * ((F[r] - F[l - 1] + mod) % mod) % mod) % mod;}printf("%lld\n", (ans % mod + mod) % mod);
}