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RSS订阅原创 5.6Student类
class Student{ private String stuno; private String name; private float math; private float english; private float computer; public Student(){} public Student(String stuno,String name,float mat
2014-06-01 08:03:36
605
原创 poj2196
#includeint calc(int k,int b){ //k的b进制 sum int sum=0; while(k){ sum+=k%b; k/=b; } return sum;}int main(){ int i=2992; for(;i<=9999;i++) if(calc(i,10)==calc(i,12)&&calc(i,12)==
2014-05-30 21:14:44
483
原创 poj1504
Adding Reversed NumbersTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 14775 Accepted: 8139DescriptionThe Antique Comedians of Malidinesia prefer come
2014-05-25 18:52:26
414
原创 hdoj1087
先存下#includeint a[1001],dp[1001];int main(){ int n,i,j,max,maxn; while(scanf("%d",&n)!=EOF&&n!=0) { for(i=1;i<=n;i++){ scanf("%d",&a[i]); dp[i]=a[i]; } maxn=-1; for(i=2;i<=n;i++
2014-05-18 08:37:51
421
原创 hdoj动态规划A
Problem ATime Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 38 Accepted Submission(s) : 16Problem Description在讲述DP算法的时候，一个经典的例子就是数塔
2014-05-05 09:36:04
503
原创 hdoj1048
#include#includechar trans(char c){ if(c>=65&&c<=69) return c+21; else if(c>69&&c<=91) return c-5; else return c;}using namespace std;int main(){ string beg,end,str; while(1) { ge
2014-05-02 10:16:02
323
原创 hdoj2026
#include#includeint main(){ char a[100]; while(gets(a))//use gets(); { int n=strlen(a); a[0]=a[0]-32; for(int i=1;i<n-1;i++) { if(a[i]==' ') a[i+1]-=32; } printf("%s\n",a); }
2014-04-26 20:47:58
465
原创 hdoj2017
#includeint main(){ int n; char ch; while(scanf("%d",&n)!=EOF&&n!=0) { getchar(); int count=0; while(n--) { while((ch=getchar())!='\n') { if(ch>='0'&&ch<='9') count++; }
2014-04-24 18:16:53
417
原创 hdoj2016
#includeint main(){ int n,a[100]; while(scanf("%d",&n)!=EOF) { if(n==0)break; for(int i=0;i<n;i++) scanf("%d",&a[i]); int min=a[0],j=0; for(int i=0;i<n;i++) if(min>a[i]) {
2014-04-24 17:57:13
431
原创 hdoj2013
#includeint main(){ int n; while(scanf("%d",&n)==1) { int sum=1; for(int i=n;i>1;i--) sum=2*(sum+1); printf("%d\n",sum); } return 0;}
2014-04-23 09:46:12
286
原创 hdoj2012
#includeint is_prime(int n){ int i; if(n<2) return 0; for(i=2;i*i<n;i++) { if(n%i==0) return 0; } return 1;}int main(){ int x,y; while(scanf("%d%d",&x,&y)==2&&(x!=0||y!=0)) { int cou
2014-04-23 09:39:11
375
原创 hdoj2011
#includevoid fun(int n){ int j,count=1,q; double sum=0; for(j=1;j<=n;j++) { if(count%2==1)q=1; else q=-1; sum+=1.0*q/j; count++; } printf("%.2lf\n",sum);}int main(){ int m,n,i; sc
2014-04-23 09:04:40
376
原创 hdoj2010
水仙花数Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 92329 Accepted Submission(s): 27357Problem Description春天是鲜花的季节，水仙花就是其中最迷人的代表，数学上有个水仙花
2014-04-23 08:54:30
399
原创 C++primer plus（第四版）第五章习题参考
本人是新手，写博客也是为了记下自己的学习过程如果有错的地方还请大家斧正转载请说明出处5.1#includeusing namespace std;int main(){ int low,high,sum=0; cin>>low>>high; for(int i=low;i<=high;i++) sum+=i; cout<<sum<<endl; return 0;}
2014-03-20 20:21:00
675
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