Dosth_Magic

#include#includeusing namespace std;const int N = 100005;int le[N], ri[N], n, m;typedef long long ll;void link(int l, int r)//链接l和r，l在左边{ le[r] = l; ri[l] = r;}int main(){ int cas =

看不下去，找点简单的，再回来写

#include#includeconst int maxd = 20;int s[1<<maxd];int main(){ int D, I, N; scanf("%d", &N); while(N--) { scanf("%d", &D); if(D == -1) break; scanf("%d", &I);

#include int main(){ int n,i,j; scanf("%d", &n); while(n--){ int k,count[10] = {0}; scanf("%d", &k); if(k == 0) count[0] = 1; else for(i=1; i<

#includeusing namespace std;//输入一个子天平，返回天平是否平衡，参数W修改为子天平的总重量bool solve(int& W){ int W1, D1, W2, D2; bool b1 = true, b2 = true; cin >> W1 >> D1 >> W2 >> D2; if(!W1) b1 = solve(W1);

#include#includeint pile[200], temp;char str[1000];void dfs(int pos){ scanf("%d", &temp); if (temp < 0) return; pile[pos] += temp; dfs(pos-1); dfs(pos+1);}int main(){

#include#includeint main(){ char a[100]; int i,j,k=0,l,num; scanf("%d",&num); while(num--) { scanf("%s", a); l = strlen(a); for(j=1; j <= l; j++)

#include#includeconst int maxn = 100000 + 5;int last, cur, next[maxn];char s[maxn];int main(){ while(scanf("%s", s+1) == 1) { int n = strlen(s+1); last = cur = 0; nex

对不定长数组的操作：push_back()在尾部添加元素pop_back()在尾部删除元素clear()清空resize()改变大小empty()测试是否为空#include#include#include#includeusing namespace std;const int maxn = 30;int n;vector pile[maxn];//每个pile

队列：queue s定义一个队列push()入队pop()出队front()取队首元素不删除#include#include#includeusing namespace std;const int maxt = 1000 + 10;int main(){ int t, kase = 0; while(scanf("%d", &t) == 1 &&

#include#include#includeusing namespace std;const int maxcol = 60;const int maxn = 100 + 5;string filenames[maxn];void print(const string& s,int len, char extra) { cout << s; for(int

#include#includeusing namespace std;const int maxn = 10000;int main(){ int n, q, x, a[maxn], kase = 0; while(scanf("%d%d", &n, &q) == 2 && n) { printf("CASE# %d:\n", ++kase);//格式

a

集合：和离散数学上的集合一样，就是只出现一次，里的元素已经从小到大排好了#include#include#include#includeusing namespace std;set dict;int main(){ string s, buf; while(cin >> s) { for(int i= 0; i < s.length(); i++)

#include#include#include#includeusing namespace std;struct Matrix{ int a,b; Matrix(int a=0, int b=0):a(a),b(b) {}}m[26];//定义了一个结构体变量，并把两个成员都初始化为0stack s;int main(){ int n; ci

这个感觉有点难，集合套在映射里，集合套在不定长数组里，主要是把集合整体理解。#include#include#include#include#include#include#include#include#includeusing namespace std;typedef set Set;//把set 等效成Set，把set 转化成Set类型map IDcache;//定义

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;#define ALL(x) x.begin(),x.end()#define INS(X) inserter(x,x.begin())int m,n;

优先队列：“急诊病人插队”priority_queue pq声明优先队列"越小的整数优先级越低的队列"push()入队pop()出队top()取队首元素这个题像了一会，纠结在为什么i == 1500时停，原因就在于这个大循环每次会把最小的那个整数提出来，制造3个新的丑数，然后出队。这样每一循环，按小到大顺序出队一个丑数，直到1500个#include#include

Background背景Bin packing, or the placement of objects of certain weights into different bins subject to certain constraints, is an historically interesting problem. Some bin packing problems are

题中输出时要求“......without trailing and leading spaces......”，否则会wrong answer.1、学会getline的一种用法。2、学会istringstream的用法。3、熟悉c++中的格式输出//#define LOCAL #include #include #include #include #i

代码非常清晰，考虑到所有的情况学习这种写法#include#includechar str[10][10], Deal[3000];char ch;int ErrorFlag;int x,y;int Flag;int main(){ Flag=0; while( (gets(str[0])) != NULL && strcmp(str[0],"Z") ){

#IDTitleSolved byDifficulty1100The 3n + 1 problem2648512102Ecological Bin Packing106082310071Back to High School Physics95543410055Hashmat t

#include#includeusing namespace std;std::map g_Tb1;int CycleLength(int n){ int &cl = g_Tb1[n]; if(cl == 0){ n = (n % 2 == 0) ? (n / 2) : (3 * n + 1); cl = 1 + CycleLength(

注释后补#include#includeconst int maxn = 100 + 5;char pic[maxn][maxn];int m, n, idx[maxn][maxn];void dfs(int r, int c, int id) { if(r = m || c = n) return ; if(idx[r][c] > 0 || pic[r][c

注释以后补！！！#include#includeconst int len = 32;const int maxn = 1024+10;char s[maxn];int buf[len][len], cnt;void draw(const char* s, int& p, int r, int c, int w) { char ch = s[p++]; if(ch

写了n遍总算是写对了，不能停止地相信自己。#include#include#include#includeusing namespace std;const int maxn = 300;struct Node{ bool have_value;//标记是否赋值 int v;//值 Node *left, *right;//左节点和右节点 Node()

#include#include#include#includeusing namespace std;const int maxv = 10000 + 10;int in_order[maxv], post_order[maxv], lch[maxv], rch[maxv];//四个数组，第一个存放中序遍历，第二个存放后序便利，后两个分别存放左节点和右节点的值int n;//n是

我知知道这题要用BFS，迷宫题，好难啊，我已无奈，苦苦坚持，~~~#include#include#include#include#includeusing namespace std;struct Node{ int r, c; int dir; Node(int rr=0, int cc=0, int ddir=0):r(rr), c(cc), dir(d

c++11CE了c++AC了#include#include#include#include#define MAXN 30using namespace std;int vis[MAXN], N, T, f[MAXN],rank[MAXN], in[MAXN], out[MAXN];void init(){ for(int i = 0; i < MAXN; i++)

搜到一个代码 我尼玛 c语言 顿时感到从来没有的亲切写着写着发现什么语言都看不懂啊 泪奔如若有人看到此篇 教导一二 或仅留几句到此一游 我会尤为感激 代码没白抄#include#includeint n, st[110], top, G[110][110], vis[110];void dfs(int u){ int v; vis[u] = 1; for(v

英语实在是不好，题意都看不懂，背单词又背不下去，那就慢慢翻译吧，反正没事干。翻译不求用词精准，句子也可能不通顺， 只是能理解了题意就好。To enable homebuyers to estimate the cost of flood insurance, a real-estate firm provides clients with the elevation of each 10-

好像是个很坑的代码#include#include#include#includeusing namespace std;string str[1200];string value;int main(){ int n,cas=0,k; while(cin >> n && n){ cout << "S-Tree #" << ++cas << ":"

感觉算法还是没明白，有种不理解的东西在，最需要的就是去大量的思考，思考简单的，由浅入深。分析：递归。注意树的前序，中序，后序遍历说的是根的访问顺序。            前根序：根，左子树，右子树；            中根序：左子树，根，右子树；            后跟序：左子树，右子树，根。            根据上面的三个式，观察上面三

题意:括号匹配, 按以下规则:1. 若字符串为空, 则正确.2. 若 A、B 皆正确, 则 AB 也正确.3. 若 A 正确, 则 (A) 与 [A] 都正确.即要求括号与中括号都要匹配.思路:使用栈, 如果当前字符与栈顶字符匹配(必须先进去的是左括号,外面的是右括号才算是匹配的):1. 当前字符为 ')', 栈顶字符为 '('; 2. 当前字符为 ']',

题目本身就是象棋里马走“日”，每次共8种走法，记录走过的格子，有限次数内能搜到目标格 ~~思考思考！！！ 动脑子 动脑子！！！#include#include#include#include#include#includeusing namespace std;int a[10][10],flag[10][10],count;int step[10][3] = { {1,2

Time limit: 3.000 seconds限时：3.000秒 Problem问题You are to simulate the playing of games of "Accordian" patience, the rules for which are as follows:模拟玩一个“手风琴”纸牌游戏，规则如下：Deal cards on

Dr. Tuple is working on the new data-mining application for Advanced Commercial Merchandise Inc. Tuple先生工作在新的数据挖掘应用对先进的商业货物。（明显句子不通。。。）One of the subroutines for this application works with two arra

简简单单 单单纯纯 我就是想坚持下去 到以后发现我竟然是这样入门的 自己是不是都会觉得很奇葩#include#include#includeusing namespace std;const int maxn = 200 + 10;int n;char buf[maxn][maxn];void dfs(int r, int c) { printf("%c(", buf[r

//超时的#include#includeusing namespace std;vector a;vector b;int n,x,y;int main(){ while( cin >> n && n ){ int t = 1; for(int i = 0; i < n; i++){ cin >> x >> y;

#include#include#include#include#include#include#includeusing namespace std;mapmaps;int p[10008][6];int main(){ int n,m; lop:while(cin>>n>>m){ string s,t; maps.clear