zju2016的博客

#include#include#include#include#include#include#include#includeusing namespace std;vector result;vector words;bool compare(string &s1,string &s2){ return s1<s2;}string small(string

#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())typedef set Set;

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int t;int main(){ int k=1; while(scanf("%d",&t)!=EOF) { map int2int; if(t==0)

#include#include#include#include#include#include#include#include#include#includeusing namespace std;const int arr[]={2,3,5};typedef long long LL;int main(){ priority_queue,greater >

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int N;void print(string &s,int len){ cout<<s; while(len>s.length()) { cout<<" ";

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int n,m;int main(){ while(scanf("%d %d",&n,&m)!=EOF) { int i,j; string temp;

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int count(int temp){ int total=1; while(temp!=1) { if(temp%2==0) temp=temp/2;

#include#include#include#include#include#include#include#include#include#includeusing namespace std;map int2int;void init(int t,stack st[]){ while(1) { int index=int2int[t]; int t

#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ map > str2set; map str2int; string s; getline(cin,

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int data[9];string ar[6]={"BCG","BGC","CBG","CGB","GBC","GCB"};int arr[6][9]={0,1,1,1

#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef struct{ double x,y,w,d,h; int id;}Build;int n;bool com

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int graph[35][35];bool start;int arr[30][10];int length[35];bool compare(int i,int

#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int i,j; string s; vector v[1010]; int stepv=0; int

#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int T; while(scanf("%d",&T)!=EOF) { while(T

#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int n; while(scanf("%d",&n)!=EOF&&n!=0) { i

#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int n; while(scanf("%d",&n)!=EOF&&n!=0) { i

#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int i,j; string s; map str2int; vector v; v

#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int T; while(scanf("%d",&T)!=EOF) { while(T

#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int T; while(scanf("%d",&T)!=EOF) { while(T

#include#include#include#include#include#include#include#include#include#includeusing namespace std;double percent[30][30][30];int path[30][30][30];int n;void Print_path(int i,int j,in

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int height[10010];int main(){ memset(height,0,sizeof(height)); int i; int left,hig

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int arr[1000010];int main(){ int N; while(scanf("%d",&N)!=EOF) { int c

#include#include#include#include#include#include#include#include#include#include#include#define ep 1e-9using namespace std;int main(){ int init_height,worker; while(scanf("%d %d",&in

#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef struct{ string author; int status;}book;vector

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int N; int arr[110][110]; int Sum[110][110]; while(scanf("%d"

这道题目难在阅读，翻译如下（第一段可以省略，从第二段开始）       订货薄会接收到一系列的消息，这些消息是订单或者是取消之前订单的请求。没有被取消的订单是激活状态的。订单类型分为购买订单和销售订单。每一个购买订单或者是销售订单有两个正数的变量size和price。订货薄保持了一系列激活状态的订单列表，并且产生quote(报价)和trades（交易）。激活状态的购买订单中价格最高的是最优购买

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ string s; vector> data; int lines = 0; int max_length = -1; while (getl

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int T, n; cin >> T; while (T--){ cin >> n; vector arr(n,0); set> da

#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int n; while (cin >> n){ if (n == 0) break; queue q; for (int i = 1;

建立字典树，保存计算的结果，便于下次查找#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;struct node{ int id; node *next[10];

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int n; while (cin >> n){ if (n == 0) break; map, int> amount;

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ set data; string s; while (cin >> s){ data.insert(s); } for

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;bool compare(pair a,pair b){ if (a.first == b.first) return a.second < b.secon

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int T; cin >> T; int n, m; while (T--){ cin >> n >> m; prio

有坑的有技巧的题目。首先来谈谈容易被坑到的几个地方：第一，记得考虑输入的时候人数为零的情况；第二，进入康复室的床位的排序，首先是按照病人从手术室出来的时间（注意是出来的时间，不能加入从手术室到康复室的路上的时间）与此时康复室的时间进行比较，如果从手术室出来的时间小于等于（注意是小于等于）康复室的可用时间，那么就为该病人分配床位，否则就不分配；第三点，从手术室出来的病人首先按照出手术室的时间进行排序

简单题，利用双端队列来模拟普通的执行队列（因为可能需要在队列头部进行插入），利用普通队列来模拟阻塞队列。题目中的样例输入是个坑：应该先给出Case的个数，再给出各个测试数据，但是题目中的样例没有给出Case的个数，仅仅是直接给出了一组测试数据，大家在解题的时候要小心一点。#include#include#include#include#include#include#include

#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef struct book{ string name; string author; bool flag;}book;bool com

简单的模拟题，已知一个输入序列：1，2，3......N，是否能够利用栈的Pop和Push操作得到目标的序列代码如下：#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int total; while (ci

简单题，利用栈来进行操作。读取指令，如果读取到的指令仅仅由一个字母组成，那么就查找我们是否已经保存了该矩阵的信息，如果保存了，那么不需要做任何的乘法操作，直接输出0，如果没有保存，那么就直接输出"error"。对于长度不为1的指令，依次读取相应的信息，如果是左括号，那么就往栈1中存入一个行列均为-1的矩阵，如果是其他的非右括号的符号，那么就存入相应的矩阵，如果读入的是右括号，那么就从栈中依次弹出矩

简单的题，要用链表，保持时间复杂度在O(n)，切记不要用数组、字符串，这道题目的时间卡的很紧，多次赋值会超时的。源代码如下：#include#include#include#include#include#include#include#include#includeusing namespace std;typedef struct node{ char data