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RSS订阅原创 河南省第三届省赛 AMAZING AUCTION
AMAZING AUCTION时间限制：3000 ms | 内存限制：65535 KB难度：4描述Recently the auction house has introduced a new type of auction, the lowest price auction. In this new system, people compete for the low
2014-05-20 17:32:05
699
原创 hdu1018
Big NumberTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24633 Accepted Submission(s): 11191Problem DescriptionIn many appli
2014-05-14 08:29:45
449
原创 hdu 1004
Hot~ 2014暑期多校联合训练——正式启动报名~ 详见“杭电ACM”微博~Let the Balloon RiseTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 69548 Accepted Sub
2014-05-13 22:46:34
562
原创 hdu1048
The Hardest Problem EverTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15093 Accepted Submission(s): 6981Problem Description
2014-05-12 21:11:14
560
原创 hdu1013
Digital RootsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46703 Accepted Submission(s): 14426Problem DescriptionThe digita
2014-05-12 20:53:05
673
原创 POJ 2739
DescriptionSome positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 5
2014-04-15 23:06:00
513
原创 POJ 3094
DescriptionA checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change, so checksums are often used
2014-04-13 17:40:57
371
原创 POJ 2707
DescriptionWhat do you do if you need to copy a 560x400mm image onto a standard sheet of US letter-size paper (which is about 216x280mm), while keeping the image as large as possible? You can rota
2014-04-13 15:28:13
724
原创 线性表顺序表插入操作题
#includeusing namespace std;#define Max 10struct Xian{ int data[Max]; int length;}Sqlist;int main(){ int i,j=0,n; cin>>Sqlist.length; for(i=0;i<Sqlist.length;i++) { Sqlist.data[i]=i;
2014-03-20 20:37:08
854
原创 指针和数组
#include#includeusing namespace std;int main(){ int a[10]; int *p; \\int * 是p的类型 p=a;\\ 数组的名字就是一个指针，赋值给p *p=10; p所指地址内容赋值 *(p+1)=20;指针+整数n含义 就是 指针p+n*siziof(int *) cout<<a[0]<<" "<<a[1]<
2014-03-19 15:16:24
375
原创 汉诺塔详解
汉诺塔一个有趣的游戏，递归算法的经典题目。题目不在赘述。思路：将n-1个圆盘（除最大的圆盘外）从柱子a移到b。将最大的圆盘从柱子a移动到柱子c。将柱子b上n-1圆盘从柱子b移动到c。 1，3 方法一致，都是利用辅助柱子移动圆盘#includevoid han(int n,char a, char b, char c){ if(n==1) pri
2014-03-13 11:57:34
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原创 线性表顺序表归并算法
#includeusing namespace std;struct SqList{ int data[50]; int length;} L1={{1,2,3,4,5,},5},L2={{5,6,7,8,9},5},L3;void Merge(SqList L1,SqList L2,SqList &L3){ int i=0,j=0,k=0; while(i<L1.len
2014-03-12 21:41:34
780
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