CSLDBLYDX的专栏

#include #include #include #include using namespace std;const int MAX = 100000+5; const int MIN = -32767;int a[MAX];int main(int argc, char *argv[]){ int t,n; scanf("%d",&t);

#include #include #include #include using namespace std;const int MAX = 100+5;struct P{ int x,y; bool operator <(const P& point) const { return x < point.x; }

#include #include #include #define FOR(i,s,t) for(int i = ( s); i < (t); i++)#define max(i,j) (i > j ? i : j)#define min(i,j) (i < j ? i : j)using namespace std;const int MAX = 20 + 3; cons

用高效的方法筛选小于n的素数

// UVA 1197 K Smallest Sums.cpp: 主项目文件。#include "stdafx.h"#include "stdio.h"#include #include using namespace std;const int MAX = 750+10;struct data{ int sum,x; data(int a, int b):sum(

#include #include "stdio.h"#include using namespace std;const int MAX = 1000 +10;struct rgst{ int q_num,t,p; bool operator <(const rgst& r)const { return t == r.t ? q_num > r

#include #include "stdio.h"#include #include "string.h"using namespace std;const int MAX = 20000 +10;struct node{ int r,dtr;};void set_r(node *ns,int u,int v){ ns[u].r = ns[v].r;

#include #include "stdio.h"#include "string"using namespace std;struct chr{ char val; chr *next ;};chr head,*end;int main(){ chr *cur; char c; int flag; while(scanf(

#include #include "stdio.h"using namespace std;const int MAX = 2000000+10;int r[MAX],id[MAX/2],size[MAX],sum[MAX],cnt;int find_r(int p){ if(r[p] == p)return p; return r[p] = find_r(

Neal is very curious about combinatorial problems, and now here comes a problem about words. Knowing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie.Since

#include #include #include using namespace std;const int PHON_LNG = 10 + 2,MAX_SIZE = 10000 * (10 + 2) + 5,NUMS = 10 + 2;int num[MAX_SIZE][NUMS],val[MAX_SIZE],flag;int size;int idx(char c)

#include #include "stdio.h"#include #include const int MAX = 2000000 + 10,MAX_R = 20+5,MAX_SUM = 1000000000 + 10;int set[MAX_R][MAX],add[MAX_R][MAX],sum[MAX_R][MAX],mmax[MAX_R][MAX],mmin[MAX_R][

City GameTime Limit: 3000MS Memory Limit: 30000KTotal Submissions: 4730 Accepted: 1876DescriptionBob is a strategy game programming specialist. In his new city buil

vector 放到循环里，否则每次排序都是在原有的数据基础上。

#include #include #include using namespace std;const int MAX_SIZE=20000;int main(){    int n,m;    int heads[MAX_SIZE],soldiers[MAX_SIZE];    while(scanf("%d%d",&n,&m)==2)   

#include #include #include using namespace std;int main(int argc, char *argv[]){ int n,m; while(scanf("%d%d",&n,&m)!= -1) { double total = 0,pos; for(int i = 1; i

#include #include #include #include using namespace std;const int MAX = 10000+5;struct Ant{ int id,pos,dirc; bool operator < (const Ant& a)const { return pos < a.pos; }};int main

本题是数学题数学题关键在于建立数学模型。    #include #include #include #include using namespace std;const int MAX = 1000000+5;long long p[MAX],c[MAX],sum;int main(int argc, char *argv[]){

#include #include #include using namespace std;long long m;long long for_ward(int n,long long k,int step){ long long pos = k; long len = 1; for(int i = 1; i < n; i++)len *= 10;

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.Follow up:Can you solve it without using extra space?  给定链表，若存在环返回环的起点，否则返回null。     貌似网易面

从floyd算法学习动态规划       2010-06-06 22:402824人阅读评论(1)收藏举报算法语言iniooc    floyd的相关概念看这里。    floyd的代码实现其实很简单：    [cpp] view plaincopyprint?void floyd()  {      for(k=0;k        f

Given a linked list, determine if it has a cycle in it.Follow up:Can you solve it without using extra space?   判断链表有环交了两次才AC,判断two==null后应该直接返回false,即two指空应当马上返回false.  /** * Definition f

[思路]就是简单的归并排序。 其实算法还可以优化，因为用链表进行归并，无法像数组那样直接取中间位置，进行遍历获取中间的话效率会降低挺多。可以定义一个全局变量notSortedStart，记录当前的无序位段的起始位，作为right的起点。当串长为1时将notSortedStart 右移，则notSortedStarted记录的是中间位置。只要保证递归的顺序是先左后右，即可保证当前节点一定

#include #include #include #include using namespace std;const int MAX = 100000 + 5; const long MAX_VAL = 1000000000 + 5;long arr[MAX]; int main(int argc, char *argv[]){ int n,s;

本题属于贪心+枚举；#include #include #include #include #include #include #include using namespace std;const int INF = 24;//int all_rots[24][6]={{3,1,0,5,4,2},{1,2,0,5,3,4},{2,4,0,

用中位数可证明正确性！  #include #include #include using namespace std;long f(long n){ return n == 1? 1: f(n/2) +1;}int main(int argc, char *argv[]){ long n; while(scanf("%ld",&n) !=

#include #include #include #include using namespace std;const int MAX_SIZE = 1000 * 22,CHARS = 256+ 2,NAME_LEN = 30 + 2;int spc[MAX_SIZE][CHARS],val[MAX_SIZE];int count;int size ,f,r;st