前言（在题解前面瞎扯的话）
这场比赛已经结束了有几天，但我各种忙，虽然AK但还是没来得及写题解。（我才不会告诉你我跑去学数据结构了）
T1 区间方差
(就不贴题好了)
首先可以推公式（我们可以知道，线段树然而并不能通过初中学过的方差公式在log(L)内求出方差）:
（s2表示方差，L表示区间长度，xi表示区间的每一项，最后一个x上画了一根线表示这些数据的平均数）
用二项式定理完全平方公式可得:
再次展开:
另外，再代入以下这个
得到了:
然后继续吧。。
然后duang地一声合并同类项，于是我们得到了:
然后可以高兴地发现，和 和 平方和都是线段树的拿手好戏，然而按照这么写估计只能过几个点。
很明显平方和肯定会超long long，难不成还要写一个高精度？
仔细读读题，然后就发现可以模一个1e9+7。然后处处取模（笑），最后就能够水掉这道水题了。
至于求逆元，可以用在线算法，扩展欧几里得、快速幂（好像是基于欧拉定理），也可以用线性推逆元预处理所有可能用到的逆元。
Code(线段树)
1 /** 2 * luogu.org 3 * Problem#2005 4 * Accepted 5 * Time:965ms / 691ms 6 * Memory:18601k / 18601k 7 */ 8 #include<iostream> 9 #include<fstream> 10 #include<sstream> 11 #include<cstdio> 12 #include<cstdlib> 13 #include<cstring> 14 #include<ctime> 15 #include<cctype> 16 #include<cmath> 17 #include<algorithm> 18 #include<stack> 19 #include<queue> 20 #include<set> 21 #include<map> 22 #include<vector> 23 using namespace std; 24 typedef bool boolean; 25 #define smin(a, b) (a) = min((a), (b)) 26 #define smax(a, b) (a) = max((a), (b)) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-' && x != -1); 32 if(x == -1) return; 33 if(x == '-'){ 34 x = getchar(); 35 aFlag = -1; 36 } 37 for(u = x - '0'; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - '0'); 38 ungetc(x, stdin); 39 u *= aFlag; 40 } 41 42 const int moder = 1000000007; 43 44 template<typename T1, typename T2> 45 class Pair{ 46 public: 47 T1 x; 48 T2 y; 49 Pair(T1 x, T2 y):x(x), y(y){ } 50 Pair operator +(Pair another){ 51 return Pair(x + another.x % moder, y + another.y % moder); 52 } 53 }; 54 55 typedef class TreeNode{ 56 public: 57 int sum; 58 int sum2; 59 int from, end; 60 TreeNode* left, *right; 61 TreeNode(long long sum, int from, int end):from(from), end(end), left(NULL), right(NULL), sum(sum){ 62 sum2 = sum * sum % moder; 63 } 64 void pushUp(){ 65 this->sum = this->left->sum + this->right->sum; 66 this->sum %= moder; 67 this->sum2 = this->left->sum2 + this->right->sum2; 68 this->sum2 %= moder; 69 } 70 }TreeNode; 71 72 typedef class SegTree{ 73 public: 74 TreeNode* root; 75 76 SegTree():root(NULL){ } 77 SegTree(int size, int* val){ 78 build(root, 1, size, val); 79 } 80 81 void build(TreeNode*& node, int from, int end, int* val){ 82 node = new TreeNode(0, from, end); 83 if(from == end){ 84 node->sum = val[from]; 85 node->sum2 = (int)(val[from] * 1LL * val[from] % moder); 86 return; 87 } 88 int mid = (from + end) >> 1; 89 build(node->left, from, mid, val); 90 build(node->right, mid + 1, end, val); 91 node->pushUp(); 92 } 93 94 void update(TreeNode*& node, int index, int val){ 95 if(node->from == index && node->end == index){ 96 node->sum = val; 97 node->sum2 = (int)(val * 1LL * val % moder); 98 return; 99 } 100 int mid = (node->from + node->end) >> 1; 101 if(index <= mid) update(node->left, index, val); 102 else update(node->right, index, val); 103 node->pushUp(); 104 } 105 106 Pair<long long, long long> query(TreeNode*& node, int from, int end){ 107 if(node->from == from && node->end == end){ 108 return Pair<long long, long long>(node->sum, node->sum2); 109 } 110 int mid = (node->from + node->end) >> 1; 111 if(end <= mid) return query(node->left, from, end); 112 if(from > mid) return query(node->right, from, end); 113 return query(node->left, from, mid) + query(node->right, mid + 1, end); 114 } 115 116 }SegTree; 117 118 /* 119 int pow_mod(int x, int pos){ 120 if(pos == 1) return x; 121 int temp = pow_mod(x, pos / 2); 122 if(pos & 1) return (int)(temp * 1LL * temp % moder * x % moder); 123 return (int)(temp * 1LL * temp % moder); 124 } 125 */ 126 127 void gcd(int a, int b, int& d, int& x, int& y){ 128 if(b == 0){ 129 d = a;x = 1;y= 0; 130 }else{ gcd(b, a % b, d, y, x); y -= x * (a / b); } 131 } 132 133 int getInv(int a, int n){ 134 int d, x, y; 135 gcd(a, n, d, x, y); 136 return (x + n) % n; 137 } 138 139 int n, m; 140 SegTree st; 141 int* initer; 142 143 inline void init(){ 144 readInteger(n); 145 readInteger(m); 146 initer = new int[(const int)(n + 1)]; 147 for(int i = 1; i <= n; i++) 148 readInteger(initer[i]); 149 st = SegTree(n, initer); 150 } 151 152 inline void solve(){ 153 int op, a, b; 154 while(m--){ 155 readInteger(op); 156 readInteger(a); 157 readInteger(b); 158 if(op == 1){ 159 st.update(st.root, a, b); 160 }else if(op == 2){ 161 Pair<long long, long long> p = st.query(st.root, a, b); 162 long long L = b - a + 1; 163 // long long inv = pow_mod(L * L % moder, moder - 2); 164 int inv = getInv(L * L % moder, moder); 165 long long fm = (L * (p.y % moder) % moder - p.x * p.x % moder + moder) % moder; 166 long long ans = fm * inv % moder; 167 printf("%d ", (int)ans); 168 } 169 } 170 } 171 172 int main(){ 173 init(); 174 solve(); 175 return 0; 176 }
不过这道题只有单点修改，那么，可以考虑用树状数组。树状数组的常数比线段树小这是家喻户晓的事情（两者理论时间复杂度一样）。
另外鉴于洛谷鬼蓄的评测机，把某些只超一点点的取模符号改成三目运算符，就快了100多ms（吐血）。
由于我比较懒，又赶去学Splay，所以没有来得急写。
T2 漂浮的鸭子
一道强连通分量的裸题。常见的算法是Tarjan:
1 /** 2 * luogu.org 3 * Problem#2049 4 * Accepted 5 * Time:150ms 6 * Memory:18527k 7 */ 8 #include<iostream> 9 #include<fstream> 10 #include<sstream> 11 #include<cstdio> 12 #include<cstdlib> 13 #include<cstring> 14 #include<ctime> 15 #include<cctype> 16 #include<cmath> 17 #include<algorithm> 18 #include<stack> 19 #include<queue> 20 #include<set> 21 #include<map> 22 #include<vector> 23 using namespace std; 24 typedef bool boolean; 25 #define smin(a, b) (a) = min((a), (b)) 26 #define smax(a, b) (a) = max((a), (b)) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-' && x != -1); 32 if(x == -1) return; 33 if(x == '-'){ 34 x = getchar(); 35 aFlag = -1; 36 } 37 for(u = x - '0'; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - '0'); 38 ungetc(x, stdin); 39 u *= aFlag; 40 } 41 42 int cv = 0, cc = 0; 43 stack<int> s; 44 int* visitID; 45 int* exitID; 46 boolean* instack; 47 boolean* visited; 48 int* belong; 49 int* next; 50 51 void getSonMap(int end){ 52 int d; 53 cc++; 54 do{ 55 d = s.top(); 56 s.pop(); 57 belong[d] = cc; 58 instack[d] = false; 59 }while(d != end); 60 } 61 62 void Tarjan(int node){ 63 visitID[node] = exitID[node] = ++cv; 64 visited[node] = true; 65 s.push(node); 66 instack[node] = true; 67 if(!visited[next[node]]){ 68 Tarjan(next[node]); 69 smin(exitID[node], exitID[next[node]]); 70 }else if(instack[next[node]]){ 71 smin(exitID[node], visitID[next[node]]); 72 } 73 if(exitID[node] == visitID[node]) 74 getSonMap(node); 75 } 76 77 int n; 78 int *len; 79 int *clen; 80 81 inline void init(){ 82 readInteger(n); 83 visitID = new int[(const int)(n + 1)]; 84 exitID = new int[(const int)(n + 1)]; 85 instack = new boolean[(const int)(n + 1)]; 86 visited = new boolean[(const int)(n + 1)]; 87 belong = new int[(const int)(n + 1)]; 88 next = new int[(const int)(n + 1)]; 89 len = new int[(const int)(n + 1)]; 90 memset(visited, false, sizeof(boolean) * (n + 1)); 91 memset(instack, false, sizeof(boolean) * (n + 1)); 92 for(int i = 1; i <= n; i++){ 93 readInteger(next[i]); 94 readInteger(len[i]); 95 } 96 } 97 98 int result = 0; 99 boolean* vis; 100 101 inline void solve(){ 102 for(int i = 1; i <= n; i++) 103 if(!visited[i]) 104 Tarjan(i); 105 delete[] visitID; 106 delete[] visited; 107 delete[] instack; 108 delete[] exitID; 109 clen = new int[(const int)(cc + 1)]; 110 vis = new boolean[(const int)(cc + 1)]; 111 memset(vis, false, sizeof(boolean) * (cc + 1)); 112 for(int i = 1; i <= n; i++){ 113 if(!vis[belong[i]]){ 114 int j = i; 115 int l = 0; 116 while(belong[next[j]] == belong[i] && next[j] != i){ 117 l += len[j]; 118 j = next[j]; 119 } 120 l += len[j]; 121 vis[belong[i]] = true; 122 smax(result, l); 123 } 124 } 125 printf("%d", result); 126 } 127 128 int main(){ 129 init(); 130 solve(); 131 return 0; 132 }
仔细读题，每个点的出度为1，那么实际上dfs就好了，用不着Tarjan，还开辣么多数组。
1 /** 2 * luogu.org 3 * Problem#2049 4 * Accepted 5 * Time:107ms 6 * Memory:17941k 7 */ 8 #include<iostream> 9 #include<fstream> 10 #include<sstream> 11 #include<cstdio> 12 #include<cstdlib> 13 #include<cstring> 14 #include<ctime> 15 #include<cctype> 16 #include<cmath> 17 #include<algorithm> 18 #include<stack> 19 #include<queue> 20 #include<set> 21 #include<map> 22 #include<vector> 23 using namespace std; 24 typedef bool boolean; 25 #define smin(a, b) (a) = min((a), (b)) 26 #define smax(a, b) (a) = max((a), (b)) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-' && x != -1); 32 if(x == -1) return; 33 if(x == '-'){ 34 x = getchar(); 35 aFlag = -1; 36 } 37 for(u = x - '0'; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - '0'); 38 ungetc(x, stdin); 39 u *= aFlag; 40 } 41 42 int n; 43 int cc = 0; 44 int *belong; 45 int *next; 46 int *len; 47 int *dist; 48 int result = 0; 49 50 void dfs(int node, int be){ 51 belong[node] = be; 52 if(belong[next[node]] == be){ 53 int dis = dist[node] - dist[next[node]] + len[node]; 54 smax(result, dis); 55 } 56 if(belong[next[node]] == 0){ 57 dist[next[node]] = dist[node] + len[node]; 58 dfs(next[node], be); 59 } 60 } 61 62 inline void init(){ 63 readInteger(n); 64 belong = new int[(const int)(n + 1)]; 65 next = new int[(const int)(n + 1)]; 66 len = new int[(const int)(n + 1)]; 67 dist = new int[(const int)(n + 1)]; 68 memset(belong, 0, sizeof(int) * (n + 1)); 69 for(int i = 1; i <= n; i++){ 70 readInteger(next[i]); 71 readInteger(len[i]); 72 } 73 } 74 75 76 inline void solve(){ 77 for(int i = 1; i <= n; i++) 78 if(belong[i] == 0){ 79 dist[i] = 0; 80 dfs(i, ++cc); 81 } 82 printf("%d", result); 83 } 84 85 int main(){ 86 init(); 87 solve(); 88 return 0; 89 }
（瞬间剪下33行！）
T3 最大差值
找到在i前面最小的一个数，边读边更新最小值，边更新结果。（通常来说不是应该放在第一题吗）
1 /** 2 * luogu.org 3 * Problem#2061 4 * Accepted 5 * Time:317ms 6 * Memory:16406k 7 */ 8 #include<iostream> 9 #include<fstream> 10 #include<sstream> 11 #include<cstdio> 12 #include<cstdlib> 13 #include<cstring> 14 #include<ctime> 15 #include<cctype> 16 #include<cmath> 17 #include<algorithm> 18 #include<stack> 19 #include<queue> 20 #include<set> 21 #include<map> 22 #include<vector> 23 using namespace std; 24 typedef bool boolean; 25 #define smin(a, b) (a) = min((a), (b)) 26 #define smax(a, b) (a) = max((a), (b)) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-' && x != -1); 32 if(x == -1) return; 33 if(x == '-'){ 34 x = getchar(); 35 aFlag = -1; 36 } 37 for(u = x - '0'; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - '0'); 38 ungetc(x, stdin); 39 u *= aFlag; 40 } 41 42 int n; 43 int minv; 44 int res = 0xffffffff; 45 46 inline void solve(){ 47 readInteger(n); 48 readInteger(minv); 49 for(int i = 1, a; i < n; i++){ 50 readInteger(a); 51 smax(res, a - minv); 52 smin(minv, a); 53 } 54 printf("%d", res); 55 } 56 57 int main(){ 58 solve(); 59 return 0; 60 }
T4 随机数生成器
我先讲我会的内容:
变形一下可得:
最后化简得到:
由于题目数据范围很大，那么只能过70%的数据，于是便有了分块打表:
1 /** 2 * luogu.org 3 * Problem#2062 4 */ 5 #include<iostream> 6 #include<fstream> 7 #include<sstream> 8 #include<cstdio> 9 #include<cstdlib> 10 #include<cstring> 11 #include<ctime> 12 #include<cctype> 13 #include<cmath> 14 #include<algorithm> 15 #include<stack> 16 #include<queue> 17 #include<set> 18 #include<map> 19 #include<vector> 20 using namespace std; 21 typedef bool boolean; 22 #define smin(a, b) (a) = min((a), (b)) 23 #define smax(a, b) (a) = max((a), (b)) 24 template<typename T> 25 inline void readInteger(T& u){ 26 char x; 27 int aFlag = 1; 28 while(!isdigit((x = getchar())) && x != '-' && x != -1); 29 if(x == -1) return; 30 if(x == '-'){ 31 x = getchar(); 32 aFlag = -1; 33 } 34 for(u = x - '0'; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - '0'); 35 ungetc(x, stdin); 36 u *= aFlag; 37 } 38 39 double E = 0; 40 double sum = 0; 41 42 int main(){ 43 freopen("a.txt", "w", stdout); 44 unsigned int jg = ((1 << 31) - 1) / 400; 45 cout << " if(a >= 1 && a < "<< jg <<") E = 0, sum = 0, q = 2;" << endl; 46 for(int i = 2; i <= jg * 400; i++){ 47 E = (sum + i) / (i - 1); 48 sum += E; 49 if(i % jg == 0){ 50 printf(" else if(a / %d == %d) E = %0.5llf, sum = %0.5llf, q = %d; ", jg, i / jg, E, sum, i); 51 } 52 } 53 return 0; 54 }
由于题目的数据范围有问题，所以第9个点TLE了。那么，我还是把官方正解贴出来（这个是我同学推出来告诉我的，鉴于我语文不好，只好贴官方正解了）
1 /** 2 * luogu.org 3 * Problem#2062 4 * Accepted 5 * Time:12ms 6 * Memory:16410k 7 */ 8 #include<iostream> 9 #include<fstream> 10 #include<sstream> 11 #include<cstdio> 12 #include<cstdlib> 13 #include<cstring> 14 #include<ctime> 15 #include<cctype> 16 #include<cmath> 17 #include<algorithm> 18 #include<stack> 19 #include<queue> 20 #include<set> 21 #include<map> 22 #include<vector> 23 using namespace std; 24 typedef bool boolean; 25 #define smin(a, b) (a) = min((a), (b)) 26 #define smax(a, b) (a) = max((a), (b)) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-' && x != -1); 32 if(x == -1) return; 33 if(x == '-'){ 34 x = getchar(); 35 aFlag = -1; 36 } 37 for(u = x - '0'; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - '0'); 38 ungetc(x, stdin); 39 u *= aFlag; 40 } 41 42 int n; 43 44 inline void init(){ 45 readInteger(n); 46 } 47 48 double sumer = 0; 49 double E = 0; 50 51 inline void solve(){ 52 if(n >= 1000000){ 53 printf("%0.5lf", 1.57721566490123286060651209 + log(n - 1)); 54 return; 55 } 56 E = 0; 57 sumer = 0; 58 for(int i = 2; i <= n; i++){ 59 E = (sumer + i) / (i - 1); 60 sumer += E; 61 } 62 printf("%0.5lf", E); 63 } 64 65 int main(){ 66 init(); 67 solve(); 68 return 0; 69 }