随笔分类 - HDOJ
摘要：I Think I Need a HouseboatTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5395 Accepted Submission(s): 1615Problem DescriptionFred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the l
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posted @ 2012-08-29 22:26
加拿大小哥哥
摘要：FatMouse and CheeseTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2812 Accepted Submission(s): 1089Problem DescriptionFatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is lab
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posted @ 2012-08-29 22:18
加拿大小哥哥
摘要：Jungle RoadsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2932 Accepted Submission(s): 2074Prob...
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posted @ 2012-08-29 21:02
加拿大小哥哥
摘要：最少拦截系统Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 40 Accepted Submission(s) : 16Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是这种导弹拦截系统有一个缺陷:虽然它的第一发炮弹能够到达任意的高度,但是以后每一发炮弹都不能超过前一发的高度.某天,雷达捕捉到
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posted @ 2012-08-29 20:19
加拿大小哥哥
摘要：FATETime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3985Accepted Submission(s): 1763Problem Descri...
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posted @ 2012-08-23 17:37
加拿大小哥哥
摘要：Piggy-BankTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5598Accepted Submission(s): 2796Problem ...
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posted @ 2012-08-22 20:55
加拿大小哥哥
摘要：Bone CollectorTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14134Accepted Submission(s): 5585Pro...
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posted @ 2012-08-22 10:37
加拿大小哥哥
摘要：View Code 1 /* 2 前缀子串能否有某个周期串重复k次，输出子串长度和最大的k,也就是最小周期情况下的k。 3 也就是说求前缀子串的最大循环节 4 方法： 遍历前缀子串，若周期存在则输出,关键在于如何求最小周期 5 */ 6 #include <iostream> 7 #include <cstdlib> 8 #include <cstring> 9 #include <string>10 using namespace std;11 int next[1000010];12 string s;13 void get()14 {15
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posted @ 2012-08-21 15:49
加拿大小哥哥
摘要：1 //仍然bfs+判重 2 #include <iostream> 3 #include <stdio.h> 4 #include <cstring> 5 #include <queue> 6 using namespace std; 7 typedef struct Node 8 { 9 int x,y; 10 int step; 11 }Node; 12 Node ch[1000]; 13 bool vis[10][10]; 14 int bfs(int r1,int c1,int r2,int c2) 15 { 16 int i,j,k,
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posted @ 2012-08-21 14:26
加拿大小哥哥
摘要：搬寝室Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9687Accepted Submission(s): 3257Problem Description搬寝室是很累的,xhd深有体会.时间追述2006年7月9号,那天xhd迫于无奈要从27号楼搬到3号楼,因为10号要封楼了.看着寝室里的n件物品,xhd开始发呆,因为n是一个小于2000的整数,实在是太多了,于是xhd决定随便搬2*k件过去就行了.但还是会很累,因为2*k也不小是一个不大于n的
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posted @ 2012-08-21 11:18
加拿大小哥哥
摘要：Hat’s WordsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3779Accepted Submission(s): 1432Problem DescriptionA hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.You are to find all the ha
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posted @ 2012-08-20 11:56
加拿大小哥哥
摘要：View Code 1 //hdu2087 2 #include <iostream> 3 #include <string> 4 #include <cstring> 5 using namespace std; 6 string str1,str2; 7 int next[1005]; 8 void get() 9 {10 int i = 0;11 int j = -1;12 memset(next,0,sizeof(next));13 next[0] = -1;//不加的话会死循环,因为 j = next[j]（j=0） 会赋一个垃圾数字 14 ...
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posted @ 2012-08-17 18:17
加拿大小哥哥
摘要：Number SequenceTime Limit: 10000/5000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6111Accepted Submission(s): 2744Problem DescriptionGiven two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 100
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posted @ 2012-08-17 14:36
加拿大小哥哥
摘要：#include<stdio.h>#include<math.h>#define PI 3.14159265int num,ans;void count_factorial(){ double t; t = (num*log(num) - num + 0.5*log(2*num*PI))/log(10); ans= (int)t+1; printf("%d\n",ans);}int main(){ int i,n; scanf("%d",&n); for( i=1 ; i<=n ; i++ ) { scanf(&qu
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posted @ 2012-08-16 21:39
加拿大小哥哥
摘要：//123*20 相当于 100*20 + 20*20+3 //常规方法N>=13就溢出 #include<stdio.h>#include<string.h>#include<stdlib.h>#define N 10000//因为每位里存储的是小于10000的数，所以缩小4倍 int vis[N];int main(){ int i,j,m; int c,temp; while(scanf("%d",&m)!=EOF) //scanf("%d",&m); { memset(vis,0,size
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posted @ 2012-08-16 21:14
加拿大小哥哥
摘要：小希的迷宫Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14515Accepted Submission(s): 4414Problem Description上次Gardon的迷宫城堡小希玩了很久（见Problem B），现在她也想设计一个迷宫让Gardon来走。但是她设计迷宫的思路不一样，首先她认为所有的通道都应该是双向连通的，就是说如果有一个通道连通了房间A和B，那么既可以通过它从房间A走到房间B，也可以通过它从房间B走到房间A，为了提
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posted @ 2012-08-16 16:34
加拿大小哥哥
摘要：不可摸数Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5296Accepted Submission(s): 1393Problem Descriptions(n)是正整数n的真因子之和，即小于n且整除n的因子和.例如s(12)=1+2+3+4+6=16.如果任何数m，s(m)都不等于n,则称n为不可摸数.Input包含多组数据，首先输入T,表示有T组数据.每组数据1行给出n(2<=n<=1000)是整数。Output如果n是不可
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posted @ 2012-08-13 21:13
加拿大小哥哥
摘要：//我发现，sqrt用c++提交会CE #include <iostream>#include <string>#include <cstring>using namespace std;int Prim( int n) { int i=2,cnt=0; while(n>0&&n>=i)//防止发生除0错误，因为可能素数还不够50 { if(n%i==0) { cnt++; while(n%i==0) //不会死循环因为n最终为1，则跳出内部while n/=i;...
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posted @ 2012-08-13 21:12
加拿大小哥哥
摘要：产生冠军Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5032Accepted Submission(s): 2413Problem Description有一群人，打乒乓球比赛，两两捉对撕杀，每两个人之间最多打一场比赛。球赛的规则如下：如果A打败了B，B又打败了C，而A与C之间没有进行过比赛，那么就认定，A一定能打败C。如果A打败了B，B又打败了C，而且，C又打败了A，那么A、B、C三者都不可能成为冠军。根据这个规则，无需循环较量，或许就能
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posted @ 2012-08-13 17:39
加拿大小哥哥
摘要：What Are You Talking AboutTime Limit: 10000/5000 MS (Java/Others)Memory Limit: 102400/204800 K (Java/Others)Total Submission(s): 7433Accepted Submission(s): 2298Problem DescriptionIgnatius is so lucky that he met a Martian yesterday. But he didn't know the language the Martians use. The Martian
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posted @ 2012-08-10 19:55
加拿大小哥哥
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