lunaArcher-优快云博客

原创浙工大姗姗杯round3 A - 数塔 HDU - 2084

数塔 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 44989 Accepted Submission(s): 26700 Problem Description 在讲述DP算法的时候，一个经典的例子就是数

2017-07-16 13:44:06 538

原创浙工大姗姗杯round2 G - Bad Luck Island CodeForces - 540D

D. Bad Luck Island time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The Bad Luck Island is inhabited by thre

2017-07-16 13:35:01 682

原创浙工大姗姗杯round2 CodeForces 116B Little Pigs and Wolves

B. Little Pigs and Wolves time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Once upon a time there were s

2017-07-16 13:09:24 592

原创浙工大姗姗杯round1 G - A Mathematical Curiosity HDU - 1017

A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 41840 Accepted Submission(s): 13449 Problem Description

2017-07-16 09:26:17 566

原创浙工大姗姗杯round2 CodeForces 103B Cthulhu

B. Cthulhu time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ...Once upon a time a man came to the sea. T

2017-07-16 09:23:51 776

原创浙工大姗姗杯round1 F - Rikka with Parenthesis II

Rikka with Parenthesis II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1545 Accepted Submission(s): 686 Problem Description A

2017-07-15 11:21:26 368

原创浙工大姗姗杯round1 E - Generating Fast UVA - 10098

UVA - 10098 - Generating Fast Generating permutation has always been an important problem in computer science. In this problem you will have to generate the permutation of a given string in asc

2017-07-15 11:05:47 1094

原创浙工大姗姗杯round1 D - Well-known Numbers

B. Well-known Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Numbers k-bonacci (k is integer, k

2017-07-15 10:59:56 437

原创浙工大姗姗杯round1 C - Hello World! UVA - 11636

When you first made the computer to print the sentence “Hello World!”, you felt so happy, not knowing how complex and interesting the world of programming and algorithm will turn out to be. Then you d

2017-07-15 10:50:49 532

原创浙工大姗姗杯round1 A - "Or" Game |CodeForces - 578B

A - "Or" Game CodeForces - 578B 题意：给n个数，再给一个x，对n个数里任意几个进行总共为k次的an*x操作，要求这n个数最后“OR”（或）结果最大。思路：有个坑点，一开始以为乘以最大的数就可以了，果然没那么简单。借别人一个反例比如 1 0 0 0 1 0 1 1 1 1 0 1 这3个数字; 假设要乘的数字x是2；只能乘1

2017-07-15 10:30:22 399

原创 Dancing link ZOJ3209

Dancing link外面的模板其实都是错的。。看了我一天纠错，其实就是link函数的D，U，L，R（上下左右）全部弄反了，不影响结果，但是会让人无法理解。另外不明白为什么非要建个结构体？？ #include #include using namespace std; int n, m, p, x1, y1, x2, y2; int shang[101000], xia[101000],

2017-04-18 14:23:24 383

原创天梯赛L2-016愿天下有情人都是失散多年的兄妹

pat并查集题目太多了。这道就用了dfs写。其实用bfs+set去重写更简单，在下比较辣鸡吧 #include #include #include #include #include #include using namespace std; int a, b, N, d; char c; struct person { char sex; int papa, mama; }Lu

2017-03-30 19:03:34 835

原创 51nod没过的那题

#include #include #include using namespace std; map > Lu; int a[100001], b[100001]; int main(){ int n; while(cin >> n){ int x = n; for(int i = 1; i <= n; i++){ c

2017-03-16 22:05:30 385

原创省赛Colorful Rainbows

#include #include #include #include #include using namespace std; struct straight{ double a, b; }Lu[6000], Hui[6000]; bool cmp(straight x, straight y){ if(x.a != y.a) return x.a <

2017-03-08 11:25:20 377

原创 prim算法学习

#include #include #include #include using namespace std; #define INF 10000000 int N, E; int sum; int town[600][1000]; void prim(){ int dis[1000], vis[1000], now = 0, Min; sum = 0; mems

2017-03-07 15:08:56 385

原创浙江第四届省赛第一题

#include #include #include #include #include using namespace std; #define inf 10000000 int n, m; int dir[4][2] = {0, -1, 0, 1, 1, 0, -1, 0}; int ans[5000000 + 5]; int mp[1000 + 5][1000 + 5]; type

2017-02-28 20:21:47 386

原创 dp矩阵连乘

#define NUM 51 int p[NUM];//p[0],p[1]代表第一个矩阵，剩下只需记录矩阵的列数，因为下一个矩阵的行数都等于上一个矩阵的列数 int m[NUM][NUM]; int s[NUM][NUM]; void MatrixChain (int n) { for (int i = 1; i <= n; i++) m[i][i] = 0; for (int r = 2;

2017-02-26 11:43:27 449

原创迪加斯特拉

#define NUM 100 #define maxint 10000 //顶点个数n，源点v，有向图的邻接矩阵为c //数组dist保存从源点v到每个顶点的最短特殊路径长度 //数组prev保存每个顶点在最短特殊路径上的前一个结点 void dijkstra(int n, int v, int dist[], int prev[], int c[][NUM]) { int i, j; bo

2017-02-25 18:31:09 944

原创 HDU 1800

代码wa了。先放着等以后找问题#include #include #include #include using namespace std; struct Node { int num; Node *Lu[10]; Node() { num = 0; for (int i = 0; i < 10; i++) Lu[i] = NULL; } }; int n; Nod

2017-02-21 11:14:50 442

原创浙工大15新生决赛改造字符串

#include #include using namespace std; int main() { string Lu, Hui; Hui = "QunHe"; while (cin >> Lu && EOF) { int n = Lu.length() - 1; int flag = 1; if (Lu[0] != 'Q' && Lu[n] != 'e') { p

2016-12-17 21:14:00 530

原创浙工大15新生决赛爆炸的男友力

#include #include using namespace std; int main() { int T, n, m; scanf("%d", &T); while (T--) { scanf("%d%d", &n, &m); int Lu[1001]; memset(Lu, 0, sizeof(Lu)); for (int i = 0; i < n; i++)

2016-12-17 20:18:34 285

原创浙工大15新生决赛找假币

#include #include using namespace std; int main() { int t; scanf("%d", &t); while (t--) { int Lu[8], Hui[8]; for (int i = 0; i < 7; i++) { cin >> Lu[i]; Hui[i] = Lu[i]; } sort(Hui,

2016-12-17 20:02:55 506

原创浙工大15新生赛 M大爷与数学题（最后一题）

#include using namespace std; int k; void dfs(int x, int y) { if (x > k || y > k) { printf("%d %d\n", x, y - x); return; } dfs(y, x + y); return; } int main() { while (scanf("%d", &k) != EOF

2016-12-16 23:19:37 420

原创浙工大15新生赛江老板A+B

#include #include #include using namespace std; const int prime[] = { 0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 }; int main() { int Lu[53],

2016-12-16 19:30:41 372

原创浙工大15新生赛数星座

#include #include struct point { int x, y; int num; }; point Lu[101][21]; char Hui, Yi[501][101]; int K; int vis[101]; bool dfs(int a, int b) { int flag; for (int i = 1; i <= K; i++) { flag =

2016-12-16 15:07:46 384

原创浙工大15新生赛 JLB A+B

#include #include #include using namespace std; int main() { string Lu, Hui; stack q; while (cin >> Lu >> Hui && EOF) { int l1 = Lu.length() - 1, l2 = Hui.length() - 1; int i, j, w = 0; fo

2016-12-16 14:10:23 479

原创浙工大15新生赛命运数

#include using namespace std; const int maxn = 51; typedef long long ll; int main() { ll Lu[maxn][maxn]; Lu[1][1] = 0; Lu[2][1] = 0; Lu[2][2] = 1; for (int i = 3; i < maxn; i++) { for (int j =

2016-12-16 09:29:08 358

原创浙工大15新生赛江老板的蛋糕

#include const int pi = 3.1415926; using namespace std; int main() { double D, A, B; while (cin >> D && D && EOF) { cin >> A >> B; B *= 5; D /= 2; if (D * D * pi - A - B >= 0) printf("HA

2016-12-16 09:22:30 466

原创浙工大15新生赛爱的密码

#include #include using namespace std; struct Hui { char s, e; }Lu[60]; int vis[60], vis2[60]; int main() { int n; while (scanf("%d", &n) != EOF) { memset(Lu, 0, sizeof(Lu)); memset(vis, 0, s

2016-12-16 09:17:10 350

原创浙工大新生赛 M大爷第一定律

#include using namespace std; int main() { int x; while (cin >> x && EOF && x) { if ((x + 1) % 10 == 0) printf("%d\n", x + 1); else printf("%d\n", x); } return 0; }

2016-12-16 08:45:39 450

原创浙工大15新生赛串

#include #include #include using namespace std; struct Hui { char Yi[11]; int flag; }Num[100000]; int main() { char Lu[100000]; int n, k; while (scanf("%d%d", &n, &k) != EOF) { for (int i =

2016-12-15 23:26:13 516

原创浙工大15新生赛变换的数组

#include using namespace std; int main() { int n; while (cin >> n && n && EOF) { int Lu[100001], Hui[100001]; for (int i = 1; i <= n; i++) { cin >> Lu[i]; } for (int i = 1; i <= n; i++)

2016-12-15 22:36:15 321

原创 UVa816 Abbott的复仇

#include #include #include #include #include using namespace std; const int maxn = 10; const char* dirs = "NESW"; //进入的四个方向; const char* turns = "FLR"; //可以转向的三种方式; const int dr[] = { -1, 0, 1,

2016-12-12 22:03:22 557

转载 UVa 10129 欧拉路

#include #include using namespace std; const int maxn = 26; int vis[maxn], l[maxn][maxn], start[maxn], endd[maxn]; void dfs(int u) { vis[u] = 1; for (int i = 0; i < maxn; i++) { if (l[maxn][maxn

2016-12-12 20:59:38 467

原创 UVa 交换学生

#include #include using namespace std; map Lu; char Hui, Yi; int main(){ int n; scanf("%d", &n); int num = 0; for (int i = 1; i <= n; i++){ cin >> Hui >> Yi; if (Lu[Yi] == Hui){ num++;

2016-11-25 20:58:51 528

原创 UVa 卡片游戏

#include #include using namespace std; queue que; int main(){ int n; scanf("%d", &n); for (int i = n; i >= 1; i--) que.push(i); while(que.size() - 1){ printf("%d ", que.front()); que.pop()

2016-11-25 20:47:05 567

原创 UVa 1594 Ducci

#include #include using namespace std; int Lu[5], flag; int main(){ for (int i = 1; i <= 4; i++){ scanf("%d", &Lu[i]); } int i; for (i = 1; i <= 1000; i++){ flag = 0; int m = Lu[1]; for

2016-11-25 20:30:50 318

原创算法竞赛入门第二版 4-1

#include #include using namespace std; char Lu, red[8][8]; int black[8][8], Hui[8][2]; int n, x, y, a, b, mark,w,e; int main(){ while (scanf("%d", &n)){ int flag = 0, e = 0; memset(red, 0, size

2016-11-22 14:32:00 300

原创算法竞赛入门经典第二版 4 - 2

因为没有题目，所以我只是按照书上描述的写的，输入输出可能有点差错 #include #include using namespace std; int Lu[10][10], n, sum = 0; //检验正方形边长1~n - 1各种可能情况下是否能构成正方形 void dfs(int x, int y){ int len = 1, flag; if (Lu[x + 1][y] !=

2016-11-22 12:18:19 401

原创 HDU ACM 16 2585 hotel

#include #include using namespace std; char hotel[100], Lu[100]; int len, len1; bool dfs(int a, int b){ if (a == len1 && b == len) return true; if (a == len1 || b == len) return false; if (ho

2016-11-19 11:04:28 372

空空如也

ACM的水题。明明运行正确，但提交就是wrong answer呐，大佬求教