StriveBen的博客

0x01 题目描述Input 多组输入，每组输入为一行字符串，例如： a < b or c < d and e < f每个符号都用空格间隔Output 假链跳到０，真链跳到１，表达式序号从100开始排。Example Input1a < b or c < d and e < fExample Output100(j<,a,b,1)101(j,_,_,102)102(j<,c,d,10

#includeint a[101000];int main(){    int n,p,i,m,k;    scanf("%d %d",&n,&p);    memset(a,-1,sizeof(a));    for(i=1;i    {        scanf("%d",&m);        if(a[m]==-1)        {   

开放定址法处理冲突其中，H(key)为哈希函数   m为哈希表长   di为增量序列可有下列3种取法：(1).   ，称为线性探测再散列；(2). 称二次探测再散列。(3).      di  =   伪随机数序列，称伪随机探测再散列。#includeint a[505];int main(){ int n,p,i,j,m,k;

#includeint a[10005];int n;void shell(int d){ int i,j,t; for(i=d;i<n;i++) { t=a[i]; for(j=i;j>=d&&a[j-d]>t;j-=d)//减到0为止 { a[j]=a[j-d]; }

用数组来建立树的模型，每次要从m/2开始；#includeint a[11];void dun(int s,int m){ int l=s; int ll=2*s; int lr=2*s+1; int h; if(s<=m/2) { if(lla[ll]) { l=ll; }

#include#includeint a[100001],t[100001];long long num;void sort(int b1,int e1,int b2,int e2){    int p=0,p1=b1,p2=b2;    while(p1    {        if(a[p1]        {            t[p++]=

#include#includeusing namespace std;typedef struct no{ char data; struct no *lc,*rc;} node;char a[50],b[50];node *creat(char *x,char *z,int k){ node *p; p=new node; if(k==0

#include#include#include#includeusing namespace std;typedef struct no{ char data; struct no *lc,*rc;} node;int i=0;char a[500],b[500];node *creat(char *a,char *b,int k){ int m;

#include#includetypedef struct no{ char data; struct no *lc,*rc;}node;int l=0,k=0;char a[50];node *creat(){ node *p; char c; c=a[l++]; if(c==',') { return N

#includechar a[]="n5!wpe^o7au";char b[]="usimdavolen";int main(){ int t; char s[101]; scanf("%d\n",&t); for(int k=1;k<=t;k++) {gets(s); int l=strlen(s); for

#include#define inf 0x3f3f3f3fint road[1005][1005];int visit[1005];int cost[1005];int n,m,total;int prime(){ int i,j,k,t,f; visit[1]=1; for(i=1;i<=n;i++) { cost[i]=road[i][1]; } f

#includeusing namespace std;typedef struct no{ int data; int step;} node;int num[2000][2000],visit[10010];int bfs(int b,int k){ node t,f; int i; queue q; t.data=b;

#includeusing namespace std;int fi[105],num;typedef struct no{ int u,v,q;}node;node a[10005];int cmp(node d,node b){ return d.q<b.q;}int find(int n){ while(n!=fi[n]) {

#includeusing namespace std;struct no{ char name[15];}a[35005][2005];char st[15];int cou[2500];int cmp(no a,no b){ return strcmp(a.name,b.name)<0;}int main(){ int n,m,i,j,k,t;

#include#includeint num[10005];int quicksort(int l,int r){ int i,j,k; if(l>r) return 0; k=num[l]; i=l; j=r; while(i<j) { while(num[j]>=k&&i<j) {

#include int a[1000005] ; int part (int *p ,int b ,int e) { int key ; key = p[b] ; while(b < e) { while (b = key) { e -- ;

0x00 前言最近在学编译原理，一个简单的词法分析程序。0x01 题目描述小C语言文法 1. <程序>→(){<声明序列><语句序列>}2. <声明序列>→<声明序列><声明语句>|<声明语句>|<空>3. <声明语句>→<标识符表>;4. <标识符表>→<标识符>,<标识符表>|<标识符>5. <语句序列>→<语句序列><语句>|<语句>6. <语句>→< if语句>|< while语

#include#includetypedef struct{ int data[10025]; int top; int base;} seqstack;seqstack s;void seqstackinit(seqstack *s){ s->top=0; s->base=0;}void seqstackpush(seqstack &s,int k){ s.data[s.top]=k; s.t

#include using namespace std;class sss{ public:int a[3]; void sort(); void display();};void sss::sort(){int t,i,j; for(i=0;i<2;i++) { for(j=0;j<3-

#include#includeint num[10005];int quicksort(int l,int r){ int i,j,k; if(l>r) return 0; k=num[l]; i=l; j=r; while(i<j) { while(num[j]>=k&&i<j) {

用优先队列做，凡是小于2的度进入队列。第一组数据有点问题。#include#include#include#includeusing namespace std;int net[10005];int main(){ int t,i,n,m,x,y,a,b,num,k; priority_queue,greater > q; scanf("%d",&t);

#include #include #include using namespace std;typedef struct bitnode{ int data; int d; struct bitnode *lchild, *rchild;}* bitree;int deep(bitree &t){ if(!t) return -1

#includeusing namespace std;int num[1000005];int erfen(int l,int r,int key){ int mid=l+(r-l)/2; if(l<=r) { if(num[mid]==key) { return mid; } else {

#includeusing namespace std;int bb[1000];priority_queue,greater > q;int main(){ int n,i,lc,m; scanf("%d",&n); memset(bb,0,sizeof(bb)); for(i=0; i<n; i++) { scanf("%d",

#includeusing namespace std;string st;int bb[1000];priority_queue,greater > q;int main(){    int i,la,lc;    while(cin>>st)    {        la=8*st.size();        memset(bb,0,sizeof(

#includeusing namespace std;typedef struct no{ char data[25]; int num; struct no *lc,*rc;}node;int n;node *creat(char *n){ node *p; p=new node; p->num=1; strcpy(p->d

#includeusing namespace std;typedef struct no{ int data; struct no *lc,*rc;}node;node *creat(int n){ node *p; p=new node; p->data=n; p->lc=NULL; p->rc=NULL; retur

#includeint a[1010];int main(){ int n,p,i,m,j,k; while(~scanf("%d %d",&n,&p)) { memset(a,-1,sizeof(a)); for(i=1;i<=n;i++) { scanf("%d",&m);

#include#define inf 0x3f3f3f3fint a[101][101];int main(){ int n,m,i,j,k,u,v,q; while(~scanf("%d %d",&n,&m)) { memset(a,0,sizeof(a)); for(j=0; j<=n; j++) {

AOE网上的关键路径Time Limit: 1000MS Memory limit: 65536K题目描述    一个无环的有向图称为无环图（Directed Acyclic Graph），简称DAG图。    AOE(Activity On Edge)网：顾名思义，用边表示活动的网，当然它也是DAG。与AOV不同，活动都表示在了边上，如下图所示：   

#include#include#includetypedef struct{    char *ch;    int length;}str;int i,j;char a[200],b[200];void strassign(str &t,char *s){    char *c;    if(t.ch)   {    t.ch=NUL

#include#include#includeusing namespace std;char a[110],b[110];int main(){    while(~scanf("%s",a))    {        scanf("%s",b);        if(strstr(a,b))            printf("YES\n");

#include#includetypedef struct{    int data[100250];    int top;    int base;} seqstack;seqstack s;void seqstackinit(seqstack *s){    s->top=0;    s->base=0;}vo

#include#include#includetypedef struct no{    char *data;    int base;    int top;}node;char ch[105];void csh(node &s){    s.data=(char *)malloc(105*sizeof(char));    s.base=

#include#include#include#define stsize 55typedef struct no{    char *data;    int base;    int top;}node;node stack1;char s[51];void csh(node &st){    st.data=(char *)mallo

链表做法#include#include#includetypedef struct no{    int data;    struct no *next;} node;node *creat(int n){    node *head,*p,*tail;    head=(node *)malloc(sizeof(node));    hea

#include#include#includetypedef struct no{    int data;    struct no *next;}node;node *creat(int n){    node *head,*p,*tail;    head=(node *)malloc(sizeof(node));    head->next

#include#include#includetypedef int elemtype;typedef struct{    elemtype *elem;    int length;    int listsize;} sq;int sqlist(sq *l){    l->elem=(elemtype *)malloc(20005*sizeo

#include#include#includetypedef int elemtype;typedef struct{    elemtype *elem;    int length;    int listsize;}sq;int sqlist(sq *l){    l->elem=(elemtype *)malloc(5*sizeof(ele

类似新建表但没有建新表。#include#include#includetypedef int elemtype;typedef struct{    elemtype *elem;    int length;    int listsize;}sq;int sqlist(sq *l){    l->elem=(elemtype *)mallo