[前言]
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\n", (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
一道强连同分量的裸题。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;\n", 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 }
时间： 2024-10-25 07:53:23