其他

題目類型：迴圈鏈表題目:一個迴圈的圈，由兩個人從頭尾開始尋找，分別將第m,第n的人刪除，（兩人是同時進行的），即是數完了才刪除的，遇到相同的只輸出一個，否則都輸出，然後重新開始數，直至全部輸完解題曆程：看題目看了很久，也許之前不仔細有陰影，這次還是看了但依然有點不是很清楚地，總而言之，有點。。。。把題目看懂，手工類比一遍，寫演算法（可惜有漏洞。。。就是沒考慮刪除一次後將指標重設時可能出現指標的下一位可能會被刪除。。。。，要同時出現最後只有兩個人的情況）心得：考慮一定要全面，

Nothing worthy to mention.Code:/************************************************************************* * Copyright (C) 2008 by liukaipeng                                      * * liukaipeng at gmail dot com                                         

We can solve this problem by simulating the calculating process of multiply two numbers on the paper, yet we've already known the product. We guess the least significant digit of one number, multiply it with the other number to get a 1 of the least

Simple problem. Just count it.Code:/************************************************************************* * Copyright (C) 2008 by liukaipeng                                      * * liukaipeng at gmail dot com                                     

Simply sort will do the job. Given to jobs, j1 and j2. which should be done first? Do j1 first will cause a total fine to j2 of j1.time * j2.fine and do j2 first will cause a total fine to j1 of j2.time * j1.fine.Code:/*******************************

I just simulate the flipping. We are not required to find the optimal solution, so I choose a way easy to understand. I find the biggest cake in the unsorted stack, and then check its position. If it is at the bottom of the stack, then nothing need

Straight forward simulating.Code:/************************************************************************* * Copyright (C) 2008 by liukaipeng                                      * * liukaipeng at gmail dot com                                       

題目類型：棧類比解題經曆：崩潰。。。。題目：（摘自星星之火的百度空間）你的任務是類比一種叫「Accordian」的紙牌遊戲，他的遊戲規則如下：

Let the required stack to be the sorted sequence, and the original stack to be the sequence to sort. A turtle is numbered by its position in the required stack: the top is 0, the second is 1, and so on. A turtle is said in reverse order if its

I just simulate the electing process. The RT is very slow.Code:/************************************************************************* * Copyright (C) 2008 by liukaipeng                                      * * liukaipeng at gmail dot com         

問題描述：找出最短路徑，加權的解法：典型的dp中#include<cstdio>#include<cstdlib>#include<cmath>#include<string>#include<vector>using namespace std;const int MAXHASHSIZE=10000003;const int MAXN=13;const int MAXS=30005;long long f[MAXS];const

 http://sd.csdn.net/a/20110808/302760.html導讀：本文是從《The Singular Secret of the Rockstar

問題描述：幾個人的01背包問題問題解法：看看147那道，就知道一樣的#include<cstdio>#include<cstdlib>#include<string>using namespace std;const int maxp=105*1000;const int maxn=1001;int v[105];long long d[maxp];int p[maxn],w[maxn];int main(){ int S,n; int

Vito should live on the street which is the median of his

題目類型：數學解題思路：本題可轉化為：求N^i=num,(N+1)^i=H;然後求等比數列的和，注意：等比因子q等於1時要分開討論。。。。本人就是卡在這裡，還有當輸入為（2*k+1）,1時輸出結果應為：0，H；（也是本人卡題原因），解題心得：      1 注意邊界考慮。。。。      2 浮點數比較：     fabs(a,b)<eps      3

問題描述：給出5種硬幣，面值分別為{1,5,10,25,50}，然後給出金額，要求能用這五種硬幣湊出金額的種類數類型：經典dp，做法：首先，d(i,s)表示用V（i），V（i+1），...V(5)這幾種硬幣湊成S的方法有多少種，那麼d(i,S)=d(i+1,S-V[i]*j)   (0<=j

Do what you are told to do.Code:/************************************************************************* * Copyright (C) 2008 by liukaipeng                                      * * liukaipeng at gmail dot com                                        

題目類型：類比題考查：資料結構題目經曆：1TLE，2Y解題思路：用鏈棧進行解答，類比全操作，即可。。。代碼：#include<stdio.h>#include<string.h>#include<malloc.h>int num;typedef struct node{ int data; struct node *next;}node,*stacklist;//鏈棧stacklist s[25];int create(stacklist

Let P denotes the prefix, T denotes the # of lost digits. We are searching for N, that the prefix of 2^N is P. We have an inequlity of    P*10^T <= 2^N < (P+1)*10^Tthus    log2(P*10^T) <= log2(2^N) < log2((P+1)*10^T),which is   

#include<cstdio>#include<cstdlib>#include<cmath>#include<string>#include<vector>using namespace std;const int MAXHASHSIZE=10000003;const int MAXN=13;const int MAXS=30005;long long f[MAXS];const int v[12]={0,5,10,20,50,10