其他

OrdersTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6402 Accepted: 4052DescriptionThe stores manager has sorted all kinds of goods in an alphabetical order of their labels. All the kinds having labels starting with the same letter are

排列Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9564 Accepted: 4223Description題目描述： 大家知道，給出正整數n，則1到n這n個數可以構成n！種排列，把這些排列按照從小到大的順序（字典順序）列出，如n=3時，列出1 2 3，1 3 2，2 1 3，2 3 1，3 1 2，3 2 1六個排列。 任務描述：

 B. Lucky Number 2time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputPetya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only

DivisibilityTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6873 Accepted: 2294DescriptionConsider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical

A Simple Problem with IntegersTime Limit: 5000MS Memory Limit: 131072KTotal Submissions: 19890 Accepted: 5238Case Time Limit: 2000MS DescriptionYou have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of

Sum of FactorialsTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 10449 Accepted: 3388DescriptionJohn von Neumann, b. Dec. 28, 1903, d. Feb. 8, 1957, was a Hungarian-American mathematician who made important contributions to the foundations

題目大姨：給出一個遞迴法則，求出給定變數a,b,c經過該遞迴後的值..法則如下 Consider a three-parameter recursive function w(a, b, c): if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns: 1 if a > 20 or b > 20 or c > 20, then w(a, b, c) returns: w(20, 20, 20) if a &

通過寫這個程式讓我更正了一個錯誤的認識,,, -1不一定是檔案的結束符!!!!(害我調了兩天..::>_<::) #include<iostream>#include<fstream>#include<conio.h>#include<math.h>#include<time.h>using namespace std;long P[600],HEAD,FILESIZE=0;char INPATH[255],OUTPATH[

題目大姨:n頭牛排隊，給定從2~n頭牛每頭牛前面比它號碼小的牛的頭數，確定這隊牛的排隊方式 Sample Input51210Sample

Cantoring AlongTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2963 Accepted: 1608DescriptionThe Cantor set was discovered by Georg Cantor. It is one of the simpler fractals. It is the result of an infinite process, so for this program,

題目大姨:開始給出了一個"自私數"的定義，這個定義是什麼呢？如果一個數不能分解為另一個數和那個數各位元字之和，它就是一個"自私數",舉個例子，比如87可以分解為75+7+5，所以87就不是"自私數".....這道題要輸出1~10000以內的"自私數"。。。。 思路：利用一個數和這個數各個數字之和除9同餘，假設t是"自私數",設它可以被n分解，n的各位和是s,那麼就有t%9=(n%9+s%9)%9,如果t%9為奇數,那麼(t%9+9)/2就等於s%9和n%9,否則t%9等於s%9和n%9,這樣就求

ParencodingsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 11181 Accepted: 6583DescriptionLet S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways: q By an integer sequence P = p1 p2...pn where pi

題目大姨：求下列序列中第i位是多少。。。序列如下11212312341234512345612345671234567812345678912345678910123456789101112345678910。。。。。這個序列規律就是1   12   123   1234   12345 .........123456789   12345678910 ...... 思路:定義一個數組，記錄每個數的長度，然後求num[1]+......+num[k]<n的最大k,再在k+1裡找那個數輸出.

 想測試下遞迴和非遞迴運行效率的差異，因此我把八皇后以同樣演算法用遞迴和非遞迴實現了一遍，測試結果如下：    問題規模較小時非遞迴效率稍高，但規模一大就完了。。  遞迴與非遞迴相比，遞迴的優點有：代碼簡潔，可閱讀性強。遞迴缺點，浪費空間。 測試的代碼，比較簡單： #include<iostream>#include<time.h>using namespace std;int s=0,n;void queen()/*n皇后非遞迴*/{int

題目大姨:一個數T可以由1 2 3 4 .......N來表示，其中兩數之間或是加號或是減號。。。比如12 = -1+2+3+4+5+6-7,,,,12最小可由1~7這7個數以上述規則得到，現在給定一個T，求按照上述規則可以表示T的最小N。。。。。Sample Input12Sample

Equation SolverTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 427 Accepted: 266DescriptionWrite a program that can solve linear equations with one variable.InputThe input will contain a number of equations, each one on a separate line.

題目大意:求C(m)(n)，輸出結果的二進位不超過32位。。思路：先將1~100的素數打表，然後把每個分子分母分解質因數，記錄下分子因數分解的情況，分子質因數表加到一起，分母的表加到一起，最後把分子的質因數表減去分母的質因數表，然後結果就好球了。。#include<iostream>#include<cmath>using namespace std;int prime[50];void find(){prime[0]=2;int

題意：求一個序列中最長上升子序列Sample Input71 7 3 5 9 4 8Sample Output4思路:以第i位元素開始的最長上升子序列長度等於1加上i+1~n中的某個j位的最長上升子序列(a[i]<a[j])，然後遞迴求解#include<iostream>using namespace std;int n,a[1010],maxl[1010];void find(int i){int j,flag=0;for(j=i+1;j<=n;j++)if(a[j]

@interface StampTransformmer+ (unsigned long) GetStamp :(NSString *) input;@end@implementation StampTransformmer+ (unsigned long) GetStamp :(NSString*) input{ const char *timestr; unsigned long ret; unsigned long lyearCount; unsigned

題目大意：給定字型大小，和一個數，將這個數按給定的字型大小顯示出來。。。。Sample Input2 123453 678900 0Sample Output -- -- -- | | | | | | | | | | | | -- -- -- -- | | | | | | | | | | -- -- -- ---