牛客多校还是挺难的。三个人各种白给。
题目链接：https://ac.nowcoder.com/acm/contest/881#question
A：
题意就是找到两个数组第一次出现极小值位置不一样的情况。拿样例来说：a={3,1,5,2,4}, b={5,2,4,3,1}。a数组在第四个位置出现了极小值，而b数组在第四个位置并没有出现极小值。
故用两个单调递增栈来维护极小值位置最方便。只要两个栈弹出数量不一样就说明出现了极小值位置不一的情况。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 const int maxn = 1e5 + 10; 22 int n, a[maxn], b[maxn]; 23 stack<int>s1, s2; 24 25 int main() { 26 a[0] = b[0] = 0; 27 while (~scanf("%d", &n)) { 28 rep1(i, 1, n) scanf("%d", &a[i]); 29 rep1(i, 1, n) scanf("%d", &b[i]); 30 while (!s1.empty()) s1.pop(); 31 while (!s2.empty()) s2.pop(); 32 s1.push(1); s2.push(1); 33 int p; 34 for (p = 2; p <= n; p++) { 35 int num = 0; 36 while (!s1.empty() && a[p] < a[s1.top()]) num++, s1.pop(); 37 while (!s2.empty() && b[p] < b[s2.top()]) num--, s2.pop(); 38 if (num) break; 39 s1.push(p); s2.push(p); 40 } 41 printf("%d ", p - 1); 42 } 43 return 0; 44 }
B：
看到一个极好的题解：https://www.cnblogs.com/Dillonh/p/11209476.html
C：
很显然是个约束优化问题，必然是拉格朗日乘数法。
但是官方题解里有个很秀的地方：跟平常的
不同，把函数phi(x,y)前面的系数λ变为2λ，从而把原式化为一个每一项为完全平方项的式子再求和。对于每一项，单独求出最值即可。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 int n, m; 22 23 int main() { 24 while (~scanf("%d%d", &n, &m)) { 25 int a[n], s1 = 0; ll s2 = 0; 26 rep0(i, 0, n) { 27 scanf("%d", &a[i]); 28 s2 += (ll)a[i] * a[i]; 29 } 30 sort(a, a + n, greater<int>()); 31 pair<ll, ll> ans(0, 1); 32 rep0(i, 0, n) { 33 s1 += a[i], s2 -= (ll)a[i] * a[i]; 34 if (i + 1 == n || (s1 - m) >= (i + 1LL)*a[i + 1]) { 35 auto tmp = mp((ll)(s1 - m) * (s1 - m) + (i + 1LL) * s2, i + 1); 36 if (tmp.first * ans.second > ans.first * tmp.second) 37 ans = tmp; 38 } 39 } 40 ans.second *= (ll)m * m; 41 auto g = __gcd(ans.first, ans.second); 42 if (ans.first % ans.second == 0) printf("%lld ", ans.first / ans.second); 43 else printf("%lld/%lld ", ans.first / g, ans.second / g); 44 } 45 return 0; 46 }
还有一个用贪心做的不错的做法：https://www.cnblogs.com/Dillonh/p/11215846.html
D：
神仙题。可以看wzk大佬的博客：https://blog.csdn.net/nickwzk/article/details/96472362
给出std。
1 #include <bits/stdc++.h> 2 3 static const int MOD = 1e9 + 7; 4 5 void dfs(std::vector<int> &count, const std::vector<int> &a, int i, int x, 6 int p) { 7 if (i < static_cast<int>(a.size())) { 8 dfs(count, a, i + 1, x, p); 9 dfs(count, a, i + 1, x ^ a[i], -p); 10 } else { 11 count[x] += p; 12 } 13 } 14 15 int main() { 16 int n, m, k; 17 while (scanf("%d%d%d", &n, &m, &k) == 3) { 18 std::vector<int> count(1 << k); 19 for (int i = 0; i < n; ++i) { 20 std::vector<int> a(m); 21 for (int j = 0; j < m; ++j) { 22 scanf("%d", &a[j]); 23 } 24 dfs(count, a, 0, 0, 1); 25 } 26 for (int i = 0; i < k; ++i) { 27 for (int msk = 0; msk < 1 << k; ++msk) { 28 if (~msk >> i & 1) { 29 auto &x0 = count[msk]; 30 auto &x1 = count[msk | 1 << i]; 31 int tmp = x0 - x1; 32 x0 += x1; 33 x1 = tmp; 34 } 35 } 36 } 37 int inv = 1 << m; 38 int result = 0; 39 int three = 1; 40 for (int msk = 0; msk < 1 << k; ++ msk) { 41 result ^= 1LL * three * (count[msk] / inv) % MOD; 42 three = 3LL * three % MOD; 43 } 44 printf("%d ", result); 45 } 46 }
E：
可以用组合数然后去重的做法。但是如何去重是个难点。在牛客看到个大佬的组合数天秀做法：
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int mod=1e9+7; 4 const int N=1e6+10; 5 typedef long long ll; 6 ll fac[N],inv_fac[N]; 7 void exgcd(ll a, int b, ll& d, ll& x, ll& y) 8 { 9 if(!b) d = a, x = 1, y = 0; 10 else exgcd(b, a%b, d, y, x), y -= x*(a/b); 11 } 12 ll C(int n,int m,int p) 13 { 14 if(m>n) return 0; 15 if(m>n-m) m=n-m; 16 ll ans=1,tmp=1,g,x,y; 17 for(int i=0;i<m;i++)ans=ans*(n-i)%p, tmp=tmp*(i+1)%p; 18 exgcd(tmp,p,g,x,y); 19 return (ans*(x%p)%p+p)%p; 20 } 21 int main() 22 { 23 int n,m; 24 while (~scanf("%d%d",&n,&m)) 25 { 26 ll tmp1=0,tmp2=0; 27 if(n!=0) tmp1=C(2*(n+m),n-1,mod); 28 if(m!=0) tmp2=C(2*(n+m),m-1,mod); 29 ll tmp=C(2*(m+n),n+m,mod); 30 ll ans=((tmp-tmp1-tmp2)%mod+mod)%mod; 31 printf("%lld ",ans); 32 } 33 return 0; 34 }
解释可以看这篇博客：https://blog.csdn.net/qq_39239844/article/details/96475068
比较正常的做法是贪心或者dp。
贪心做法：如果字符串中只有n个AB，合法情况是每个B前面要有至少1个A；若除了AB之外还有m个BA，那每个B前面可以有超过n个A，但第一个B前面必须有至少一个A；B与后面A构成BA的时候，相当于A后面的第一个B前面少了一个A。这种做法不是很好写。
dp做法：设二维数组dp[i][j]，第一维是A的个数，第二维是B的个数。dp[i][j]代表合法方案。dp[0][0]=1。
直觉有dp[i][j]=dp[i-1][j]+dp[i][j-1]，但是要对i和j加以限制。只有 i < n || (i - n) < j 的情况才可以放A，只有j < m || (j - m ) < i 的情况才可以放B。这个条件可以从上面的贪心得到。
答案自然就是dp[n+m][n+m]。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 const int maxn = 2e3 + 10; 22 const int mod = 1e9 + 7; 23 int n, m; 24 ll dp[maxn][maxn]; 25 26 int main() { 27 while (~scanf("%d%d", &n, &m)) { 28 // init 29 rep1(i, 0, n + m) { 30 rep1(j, 0, n + m) { 31 dp[i][j] = 0; 32 } 33 } 34 dp[0][0] = 1; 35 rep1(i, 0, n + m) { 36 rep1(j, 0, n + m) { 37 if (i < n || (i - n) < j) // put A 38 dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % mod; 39 if (j < m || (j - m) < i) // put B 40 dp[i][j + 1] = (dp[i][j + 1] + dp[i][j]) % mod; 41 } 42 } 43 printf("%lld ", dp[n + m][n + m]); 44 } 45 return 0; 46 }
F：
样例给了三组毫无作用的数据，恶意满满。
最快的做法是设一个大的三角形(0,0),(10000,0),(0,10000)，枚举三角形内所有的整点并维护最终答案。然后计算答案和三角形面积比例，算出来正好逼近11/2。
实际上11/2这个数很难看出来。计算答案和三角形面积的两倍(设为S。这个用叉乘就可以做到，不仅方便而且比海伦公式精度高，海伦公式会爆long long)的比例，得到21.×××××，再试一下输出10.5S和11S，而答案正好就是11S。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 ll p1, q1, p2, q2, p3, q3; 22 23 int main() { 24 while (~scanf("%lld%lld%lld%lld%lld%lld", &p1, &q1, &p2, &q2, &p3, &q3)) { 25 pair<ll, ll> x = mp(p2 - p1, q2 - q1), y = mp(p3 - p1, q3 - q1); 26 ll ans = x.first * y.second - y.first * x.second; 27 printf("%lld ", (ll)abs(ans) * 11); 28 } 29 return 0; 30 }
G：
神仙题。
1 #include <bits/stdc++.h> 2 3 struct Solver { 4 using Hash = uint64_t; 5 static const Hash MAGIC = 1e6 + 3; 6 7 Solver(int n) 8 : n(n), bsize(get_sqrt(n)), event(n, -1), power(bsize + 1), 9 hashes(n + 1, std::vector<Hash *>(bsize)) { 10 std::vector<int> a(n); 11 for (int i = 0; i < n; ++i) { 12 scanf("%d", &a[i]); 13 } 14 power[0] = 1; 15 for (int i = 1; i <= bsize; ++i) { 16 power[i] = power[i - 1] * MAGIC; 17 } 18 blocks.emplace_back(bsize + 1); 19 for (int i = 0; i < bsize; ++i) { 20 hashes[n][i] = blocks.back().data(); 21 } 22 { 23 blocks.emplace_back(bsize + 1); 24 auto block = blocks.back().data(); 25 block[n % bsize] = -1; 26 for (int i = bsize - 1; i >= 0; --i) { 27 block[i] += block[i + 1] * MAGIC; 28 } 29 hashes[n][n / bsize] = block; 30 } 31 { 32 std::vector<int> next(n + 1, n); 33 for (int i = n - 1; i >= 0; --i) { 34 hashes[i] = hashes[i + 1]; 35 int j = next[a[i]]; 36 next[a[i]] = i; 37 if (j < n) { 38 event[j] = i; 39 blocks.emplace_back(bsize + 1); 40 int block_id = j / bsize; 41 int base = block_id * bsize; 42 auto block = blocks.back().data(); 43 Hash hh = 0; 44 for (int k = bsize - 1; k >= 0; --k) { 45 hh = hh * MAGIC + char_at(i, base + k); 46 block[k] = hh; 47 } 48 hashes[i][block_id] = block; 49 } 50 } 51 } 52 53 std::vector<int> order(n); 54 std::iota(order.begin(), order.end(), 0); 55 std::sort(order.begin(), order.end(), [this](int x, int y) { 56 int length = get_lcp(x, y); 57 return char_at(x, x + length) < char_at(y, y + length); 58 }); 59 long long result = n * (n + 1LL) / 2; 60 for (int i = 1; i < n; ++i) { 61 result -= get_lcp(order[i - 1], order[i]); 62 } 63 std::cout << result << std::endl; 64 } 65 66 static inline int get_sqrt(int n) { 67 int b = 0; 68 while (b * b <= n) { 69 b++; 70 } 71 return b; 72 } 73 74 int char_at(int s, int i) const { 75 if (i >= n) { 76 return -1; 77 } 78 return s <= event[i] ? i - event[i] : 0; 79 } 80 81 int get_LCP(int x, int y) const { 82 if (x == y) { 83 return n - x; 84 } 85 int length = 0; 86 while (char_at(x, x + length) == char_at(y, y + length)) { 87 length++; 88 } 89 return length; 90 } 91 92 int get_lcp(int x, int y) const { 93 int length = 0; 94 if (x % bsize > y % bsize) { 95 std::swap(x, y); 96 } 97 auto &hashx = hashes[x]; 98 auto &hashy = hashes[y]; 99 int xi = x / bsize; 100 int yi = y / bsize; 101 int xr = x % bsize; 102 int yr = y % bsize; 103 if (xr == yr) { 104 if (hashx[xi][xr] == hashy[yi][yr]) { 105 length += bsize - xr; 106 xi++, yi++; 107 xr = yr = 0; 108 while (hashx[xi][0] == hashy[yi][0]) { 109 xi++, yi++; 110 length += bsize; 111 } 112 } 113 } else { 114 while (true) { 115 if (hashx[xi][xr] - power[bsize - yr] * hashx[xi][xr + bsize - yr] == 116 hashy[yi][yr]) { 117 xr += bsize - yr; 118 length += bsize - yr; 119 yi++, yr = 0; 120 } else { 121 break; 122 } 123 if (hashy[yi][yr] - power[bsize - xr] * hashy[yi][yr + bsize - xr] == 124 hashx[xi][xr]) { 125 yr += bsize - xr; 126 length += bsize - xr; 127 xi++, xr = 0; 128 } else { 129 break; 130 } 131 } 132 } 133 int xx = xi * bsize + xr; 134 int yy = yi * bsize + yr; 135 while (char_at(x, xx) == char_at(y, yy)) { 136 xx++, yy++, length++; 137 } 138 return length; 139 } 140 141 int n, bsize; 142 std::vector<int> event; 143 std::vector<Hash> power; 144 std::vector<std::vector<Hash *>> hashes; 145 std::vector<std::vector<Hash>> blocks; 146 }; 147 148 int main() { 149 int n; 150 while (scanf("%d", &n) == 1) { 151 Solver{n}; 152 } 153 }
H：
线性基。
1 #include <bits/stdc++.h> 2 3 using Long = long long; 4 5 static const int D = 60; 6 static const int MOD = 1e9 + 7; 7 8 void add(int &x, int a) { 9 x += a; 10 if (x >= MOD) { 11 x -= MOD; 12 } 13 } 14 15 struct Basis { 16 Basis() 17 { 18 std::fill(index.begin(), index.end(), -1); 19 } 20 21 bool update(Long x, int i) { 22 for (int j = 0; j < D; ++j) { 23 if (~index[j] && (x >> j & 1)) { 24 x ^= basis[j]; 25 } 26 } 27 if (!x) { 28 return true; 29 } 30 int j = __builtin_ctzll(x); 31 assert(index[j] == -1); 32 index[j] = i; 33 basis[j] = x; 34 return false; 35 } 36 37 std::array<int, D> index; 38 std::array<Long, D> basis; 39 }; 40 41 int main() { 42 int n; 43 while (scanf("%d", &n) == 1) { 44 std::vector<Long> a(n); 45 for (int i = 0; i < n; ++i) { 46 scanf("%lld", &a[i]); 47 } 48 Basis o; 49 int nullity = 0; 50 int two = 5e8 + 4; 51 for (int i = 0; i < n; ++i) { 52 if (o.update(a[i], i)) { 53 nullity++; 54 add(two, two); 55 } 56 } 57 int result = 1LL * nullity * two % MOD; 58 std::vector<Long> aa; 59 for (int i : o.index) { 60 if (~i) { 61 aa.push_back(a[i]); 62 a[i] = 0; 63 } 64 } 65 Basis b; 66 for (int i = 0; i < n; ++i) { 67 b.update(a[i], i); 68 } 69 for (int i = 0; i < static_cast<int>(aa.size()); ++i) { 70 Basis bb = b; 71 for (int j = 0; j < static_cast<int>(aa.size()); ++j) { 72 if (i != j) { 73 bb.update(aa[j], j); 74 } 75 } 76 if (bb.update(aa[i], i)) { 77 add(result, two); 78 } 79 } 80 printf("%d ", result); 81 } 82 }
I：
二维平面上每个点有两个属性a和b，你需要将这些点划分成两个集合A,B，其中A中没有点在B中任意一个点的右下方（非严格），贡献定义为A中点的a属性和B中点的b属性和的和，求最大贡献。
首先把点排序(x从小到大，y从大到小)之后离散化。划分完之后，一定存在一条单调不下降的折线使得A中的点都在其上方，B中的点在线上或者下方。
考虑dp。dp[i][j]代表前i个点，j为第i个点所在横坐标的纵坐标分割线(即折线经过点(xi,j))的最大贡献。因为这条直线单调不下降，所以满足：
dp[i+1][j]j=yi=maxk<=j(dp[i][k])+bi
dp[i+1][j]1<=j<yi=dp[i][j]+ai
dp[i+1][j]yi<j<=tot=dp[i][j]+bi
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 #define mid (curl+curr>>1) 17 /* namespace */ 18 using namespace std; 19 /* header end */ 20 21 const int maxn = 1e5 + 10; 22 23 struct Point { 24 int x, y, a, b; 25 bool operator<(const Point &rhs) const { 26 if (x == rhs.x) return y > rhs.y; 27 return x < rhs.x; 28 } 29 } p[maxn]; 30 int n, b[maxn]; 31 ll lazy[maxn << 2], segt[maxn << 2]; 32 33 void pushup(int curpos) { 34 segt[curpos] = max(segt[lson], segt[rson]); 35 } 36 37 void pushdown(int curpos) { 38 if (lazy[curpos]) { 39 lazy[lson] += lazy[curpos]; 40 segt[lson] += lazy[curpos]; 41 lazy[rson] += lazy[curpos]; 42 segt[rson] += lazy[curpos]; 43 lazy[curpos] = 0; 44 } 45 } 46 47 void build(int curpos, int curl, int curr) { 48 segt[curpos] = lazy[curpos] = 0; 49 if (curl == curr) 50 return; 51 build(lson, curl, mid); 52 build(rson, mid + 1, curr); 53 } 54 55 void update(int curpos, int curl, int curr, int pos, ll val) { 56 if (curl == curr) { 57 segt[curpos] = val; 58 return; 59 } 60 pushdown(curpos); 61 if (pos <= mid) 62 update(lson, curl, mid, pos, val); 63 else 64 update(rson, mid + 1, curr, pos, val); 65 pushup(curpos); 66 } 67 68 void update(int curpos, int curl, int curr, int ql, int qr, ll val) { 69 if (ql <= curl && curr <= qr) { 70 segt[curpos] += val; 71 lazy[curpos] += val; 72 return; 73 } 74 pushdown(curpos); 75 if (ql <= mid) 76 update(lson, curl, mid, ql, qr, val); 77 if (qr > mid) 78 update(rson, mid + 1, curr, ql, qr, val); 79 pushup(curpos); 80 } 81 82 ll query(int curpos, int curl, int curr, int ql, int qr) { 83 if (ql <= curl && curr <= qr) 84 return segt[curpos]; 85 pushdown(curpos); 86 ll ans = -1e18; 87 if (ql <= mid) 88 ans = max(ans, query(lson, curl, mid, ql, qr)); 89 if (qr > mid) 90 ans = max(ans, query(rson, mid + 1, curr, ql, qr)); 91 return ans; 92 } 93 94 int main() { 95 while (~scanf("%d", &n)) { 96 rep1(i, 1, n) { 97 scanf("%d%d%d%d", &p[i].x, &p[i].y, &p[i].a, &p[i].b); 98 b[i] = p[i].y; 99 } 100 sot(p, n); 101 b[n + 1] = 0; 102 sot(b, n + 1); 103 int m = unique(b + 1, b + n + 2) - b - 1; 104 build(1, 1, m); 105 rep1(i, 1, n) { 106 int y = lower_bound(b + 1, b + m + 1, p[i].y) - b; 107 ll r = query(1, 1, m, 1, y); 108 update(1, 1, m, y, r + p[i].b); 109 if (y > 1) { 110 update(1, 1, m, 1, y - 1, p[i].a); 111 } 112 if (y < m) { 113 update(1, 1, m, y + 1, m, p[i].b); 114 } 115 } 116 printf("%lld ", query(1, 1, m, 1, m)); 117 } 118 return 0; 119 }
J：
签到题。比较两个分数的大小。Python题，Java题，C++用__int128也可以过。
1 import sys 2 3 for line in sys.stdin: 4 (x, y, a, b) = line.split() 5 x = int(x) 6 y = int(y) 7 a = int(a) 8 b = int(b) 9 if (x * b == a * y): 10 print("=") 11 elif (x * b < a * y): 12 print("<") 13 else: 14 print(">")