感觉现在Edu场比以前的难多了……
A：
温暖人心
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define init(a,b) fill(begin(a),end(a),b) 28 #define sot(a,b) sort(a+1,a+1+b) 29 #define rep1(i,a,b) for(int i=a;i<=b;++i) 30 #define rep0(i,a,b) for(int i=a;i<b;++i) 31 #define repa(i,a) for(auto &i:a) 32 #define eps 1e-8 33 #define int_inf 0x3f3f3f3f 34 #define ll_inf 0x7f7f7f7f7f7f7f7f 35 #define lson curPos<<1 36 #define rson curPos<<1|1 37 /* namespace */ 38 using namespace std; 39 /* header end */ 40 41 const int maxn = 3e5 + 10; 42 int n; 43 char s[maxn]; 44 45 int main() 46 { 47 scanf("%d", &n); 48 scanf("%s", s + 1); 49 rep0(i, 1, n) 50 { 51 if (s[i] > s[i + 1]) 52 return cout << "YES" << endl << i << " " << i + 1, 0; 53 } 54 puts("NO"); 55 return 0; 56 }
B：
数前n-10个数里几个8几个非8就完事了，正确性显然
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define init(a,b) fill(begin(a),end(a),b) 28 #define sot(a,b) sort(a+1,a+1+b) 29 #define rep1(i,a,b) for(int i=a;i<=b;++i) 30 #define rep0(i,a,b) for(int i=a;i<b;++i) 31 #define repa(i,a) for(auto &i:a) 32 #define eps 1e-8 33 #define int_inf 0x3f3f3f3f 34 #define ll_inf 0x7f7f7f7f7f7f7f7f 35 #define lson curPos<<1 36 #define rson curPos<<1|1 37 /* namespace */ 38 using namespace std; 39 /* header end */ 40 41 const int maxn = 1e5 + 10; 42 char s[maxn]; 43 int n, numOfEight = 0, re = 0; 44 45 int main() 46 { 47 scanf("%d", &n); 48 scanf("%s", s + 1); 49 rep1(i, 1, n - 10) 50 if (s[i] == '8') numOfEight++; else re++; 51 if (numOfEight > re) puts("YES"); else puts("NO"); 52 return 0; 53 }
C：
计算所有时间区间gcd，找是否存在p能整除gcd，正确性显然
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define init(a,b) fill(begin(a),end(a),b) 28 #define sot(a,b) sort(a+1,a+1+b) 29 #define rep1(i,a,b) for(int i=a;i<=b;++i) 30 #define rep0(i,a,b) for(int i=a;i<b;++i) 31 #define repa(i,a) for(auto &i:a) 32 #define eps 1e-8 33 #define int_inf 0x3f3f3f3f 34 #define ll_inf 0x7f7f7f7f7f7f7f7f 35 #define lson curPos<<1 36 #define rson curPos<<1|1 37 /* namespace */ 38 using namespace std; 39 /* header end */ 40 41 const int maxn = 3e5 + 10; 42 int n, m; 43 ll s, currX, lastX, currP, ansPos = 0, gcd; 44 45 int main() 46 { 47 scanf("%d%d", &n, &m); 48 rep1(i, 1, n) 49 { 50 scanf("%lld", &currX); 51 if (i == 1) s = currX; 52 else if (i == 2) gcd = currX - lastX; 53 else gcd = __gcd(gcd, currX - lastX); 54 lastX = currX; 55 } 56 rep1(i, 1, m) 57 { 58 scanf("%lld", &currP); 59 if (!(gcd % currP)) ansPos = i; 60 } 61 if (!ansPos) puts("NO"); 62 else printf("YES %lld %lld ", s, ansPos); 63 return 0; 64 }
D：
dp，然而我想的是数据结构……
dp做法是：定义一维数组f[maxn]，f[i]代表从当前位置往后取能取到的最大区间和；定义二维数组dp[2][maxn]，dp[0][i]表示从当前位置往前取能取到的最大区间和，dp[1][i]表示在乘x的情况下，从当前位置能取到的最大区间和。转移显而易见。
有个坑点是ans初始化应该是0而不是LONG_LONG_MIN
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define init(a,b) fill(begin(a),end(a),b) 28 #define sot(a,b) sort(a+1,a+1+b) 29 #define rep1(i,a,b) for(int i=a;i<=b;++i) 30 #define rep0(i,a,b) for(int i=a;i<b;++i) 31 #define repa(i,a) for(auto &i:a) 32 #define eps 1e-8 33 #define int_inf 0x3f3f3f3f 34 #define ll_inf 0x7f7f7f7f7f7f7f7f 35 #define lson curPos<<1 36 #define rson curPos<<1|1 37 /* namespace */ 38 using namespace std; 39 /* header end */ 40 41 const int maxn = 3e5 + 10; 42 ll a[maxn], f[maxn], dp[2][maxn], x, ans = 0; 43 int n; 44 45 int main() 46 { 47 scanf("%d%lld", &n, &x); 48 rep1(i, 1, n) scanf("%lld", &a[i]); 49 f[n + 1] = 0; 50 for (int i = n; i >= 1; i--) 51 f[i] = max(a[i], f[i + 1] + a[i]); 52 rep1(i, 1, n) 53 { 54 dp[0][i] = max(a[i], dp[0][i - 1] + a[i]); 55 dp[1][i] = a[i] * x; 56 if (max(dp[0][i - 1], dp[1][i - 1]) > 0) 57 dp[1][i] += max(dp[0][i - 1], dp[1][i - 1]); 58 ans = max(ans, max(dp[0][i], dp[1][i] + f[i + 1])); 59 ans = max(ans, dp[1][i]); 60 } 61 printf("%lld ", ans); 62 return 0; 63 }
E：
edu都有交互了，简直丧心病狂。看到形如a0+a1*x+a2*x^2+...+an*x^n的多项式就很自然会想到拉格朗日插值法。
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 #include <complex> 12 /* STL */ 13 #include <vector> 14 #include <set> 15 #include <map> 16 #include <queue> 17 #include <stack> 18 #include <algorithm> 19 #include <array> 20 #include <iterator> 21 /* define */ 22 #define ll long long 23 #define dou double 24 #define pb emplace_back 25 #define mp make_pair 26 #define fir first 27 #define sec second 28 #define init(a,b) fill(begin(a),end(a),b) 29 #define sot(a,b) sort(a+1,a+1+b) 30 #define rep1(i,a,b) for(int i=a;i<=b;++i) 31 #define rep0(i,a,b) for(int i=a;i<b;++i) 32 #define repa(i,a) for(auto &i:a) 33 #define eps 1e-8 34 #define int_inf 0x3f3f3f3f 35 #define ll_inf 0x7f7f7f7f7f7f7f7f 36 #define lson curPos<<1 37 #define rson curPos<<1|1 38 /* namespace */ 39 using namespace std; 40 /* header end */ 41 42 const int MOD = 1e6 + 3; 43 44 //定义mod P意义下的运算 45 #include <unordered_map> 46 struct Mod 47 { 48 const ll mod, sqrtMod; 49 Mod(ll p): mod(p), sqrtMod(sqrt(p) + 0.5) {} 50 51 ll add(ll a, ll b)const 52 { 53 return ((a + b) % mod + mod) % mod; 54 } 55 56 ll mul(ll a, ll b)const 57 { 58 while (a < 0) a += mod; while (b < 0) b += mod; 59 return a * b % mod; 60 } 61 62 ll pow(ll a, ll b)const 63 { 64 ll ret = 1; 65 for (a %= mod; b; b >>= 1, a = mul(a, a)) 66 if (b & 1) 67 ret = mul(ret, a); 68 return ret; 69 } 70 71 //乘法逆元 72 ll inv(ll a)const 73 { 74 while (a < 0) a += mod; 75 return pow(a, mod - 2); 76 } 77 78 ll log(ll a, ll b)const 79 { 80 unordered_map<ll, ll>x; 81 for (ll i = 0, e = 1; i <= sqrtMod; i++, e = mul(e, a)) 82 { 83 if (!x.count(e)) 84 x[e] = i; 85 } 86 for (ll i = 0, v = inv(pow(a, sqrtMod)); i <= sqrtMod; i++, b = mul(b, v)) 87 { 88 if (x.count(b)) 89 return i * sqrtMod + x[b]; 90 } 91 return -1; 92 } 93 } m(MOD); 94 95 //拉格朗日插值法 96 #define X real() 97 #define Y imag() 98 class Lagrange 99 { 100 public: 101 static vector<ll> solve(vector<complex<ll>> p) 102 { 103 vector<ll> ret(p.size()), sum(p.size()); 104 ret[0] = p[0].Y, sum[0] = 1; 105 rep0(i, 1, p.size()) 106 { 107 for (int j = p.size() - 1; j >= i; j--) 108 p[j].imag(m.mul(p[j].Y - p[j - 1].Y, m.inv(p[j].X - p[j - i].X))); 109 for (int j = i; ~j; j--) 110 { 111 sum[j] = m.add(j ? sum[j - 1] : 0, -m.mul(sum[j], p[i - 1].X)); 112 ret[j] = m.add(ret[j], m.mul(sum[j], p[i].Y)); 113 } 114 } 115 return ret; 116 } 117 }; 118 119 int main() 120 { 121 vector<complex<ll>>v; 122 for (ll i = 0, x; i < 11; i++) 123 { 124 printf("? %lld ", i); 125 fflush(stdout); 126 scanf("%lld", &x); 127 if (!x) 128 return printf("! %lld ", i), 0; 129 v.push_back({i, x % m.mod}); 130 } 131 vector<ll> ret = Lagrange().solve(v); 132 for (ll i = 0; i < m.mod; i++) 133 { 134 ll sum = 0; 135 for (ll j = 0, x = 1; j < ret.size(); j++, x = m.mul(x, i)) 136 sum = m.add(sum, m.mul(x, ret[j])); 137 if (sum == 0) 138 return printf("! %lld ", i), 0; 139 } 140 puts("! -1"); 141 return 0; 142 }
F：
给定一个无向图，满足边双联通。要求删去尽量多的边，使得剩下的子图在包含原图所有节点的情况下仍然满足边双。输出子图。
这里给个神仙用tarjan的做法，从67行开始就有点玄幻了……
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 6 const int MAXN = 10; 7 8 struct node 9 { 10 int v, next, use; 11 } edge[MAXN << 2]; 12 13 bool bridge[MAXN]; 14 int low[MAXN], dfn[MAXN], vis[MAXN]; 15 int head[MAXN], pre[MAXN], ip, sol, Count, u[MAXN], v[MAXN], n, m, p[MAXN], i, viss[MAXN], s, ans = -1, j, t[MAXN]; 16 17 void init(void) 18 { 19 memset(head, -1, sizeof(head)); 20 memset(vis, false, sizeof(vis)); 21 memset(bridge, false, sizeof(bridge)); 22 Count = sol = ip = 0; 23 } 24 25 void addEdge(int u, int v) 26 { 27 edge[ip].v = v; 28 edge[ip].use = 0; 29 edge[ip].next = head[u]; 30 head[u] = ip++; 31 } 32 bool flag; 33 void tarjan(int u) 34 { 35 vis[u] = 1; 36 dfn[u] = low[u] = Count++; 37 for (int i = head[u]; i != -1; i = edge[i].next) 38 { 39 if (!edge[i].use) 40 { 41 edge[i].use = edge[i ^ 1].use = 1; 42 int v = edge[i].v; 43 if (!vis[v]) 44 { 45 pre[v] = u; 46 tarjan(v); 47 low[u] = min(low[u], low[v]); 48 if (dfn[u] < low[v]) 49 { 50 sol++; 51 flag = 0; 52 } 53 } 54 else if (vis[v] == 1) 55 { 56 low[u] = min(low[u], dfn[v]); 57 } 58 } 59 } 60 vis[u] = 2; 61 } 62 int main(void) 63 { 64 scanf("%d %d", &n, &m); 65 for (i = 1; i <= m; i++) scanf("%d %d", &u[i], &v[i]); 66 for (i = 1; i <= m; i++) p[i] = i; 67 for (int jo = 1; jo <= 20000; jo++) 68 { 69 //随机建图 70 for (i = 1; i <= 100; i++) 71 { 72 int x = rand() % m + 1, y = rand() % m + 1; 73 swap(p[x], p[y]); 74 } 75 memset(viss, 0, sizeof(viss)); 76 for (i = 1; i <= m; i++) 77 { 78 init(); 79 viss[i] = 1; 80 for (j = 1; j <= m; j++) 81 if (!viss[j]) 82 { 83 addEdge(u[p[j]], v[p[j]]); 84 addEdge(v[p[j]], u[p[j]]); 85 } 86 flag = 1; 87 tarjan(1); 88 for (j = 1; j <= n; j++) 89 if (vis[j] == 0) 90 flag = 0; 91 if (!flag) 92 viss[i] = 0; 93 } 94 s = 0; 95 for (i = 1; i <= m; i++) 96 if (viss[i] == 1) 97 s++; 98 if (ans < s) 99 { 100 ans = s; 101 int y = 0; 102 for (i = 1; i <= m; i++) 103 if (viss[i] == 0) 104 t[++y] = p[i]; 105 } 106 } 107 cout << m - ans << endl; 108 for (i = 1; i <= m - ans; i++) 109 printf("%d %d ", u[t[i]], v[t[i]]); 110 return 0; 111 }