better_space的博客

A+B for Input-Output Practice (VII)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 63763    Accepted Submission(s): 42644Problem Des

数据的交换输出Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 88860    Accepted Submission(s): 33430Problem Description输入n(n 

字符串统计Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 68606    Accepted Submission(s): 37673Problem Description对于给定的一个字符串，统计其中数字字

偶数求和Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 75000    Accepted Submission(s): 32029Problem Description有一个长度为n(n 

数列有序!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 74766    Accepted Submission(s): 31745Problem Description有n(n Inp

绝对值排序Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 66684    Accepted Submission(s): 32228Problem Description输入n(n In

母牛的故事Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 68745    Accepted Submission(s): 34397Problem Description有一头母牛，它每年年初生一头小母牛。

发工资咯：）Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51230    Accepted Submission(s): 27970Problem Description作为杭电的老师，最盼望的日子就是每

首字母变大写Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 50133    Accepted Submission(s): 27724Problem Description输入一个英文句子，将每个单词的第一

A+B for Input-Output Practice (I)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 119606    Accepted Submission(s): 65275Problem Desc

A+B for Input-Output Practice (II)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 90028    Accepted Submission(s): 58062Problem Desc

A+B for Input-Output Practice (III)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 100794    Accepted Submission(s): 53105Problem De

A+B for Input-Output Practice (IV)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 92514    Accepted Submission(s): 49333Problem Desc

A+B for Input-Output Practice (V)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 68829    Accepted Submission(s): 46175Problem Descr

A+B for Input-Output Practice (VI)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 64447    Accepted Submission(s): 43370Problem Desc

青年歌手大奖赛_评委会打分Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 74352    Accepted Submission(s): 36954Problem Description青年歌手大奖赛中，评

A+B for Input-Output Practice (VIII)Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 138701    Accepted Submission(s): 42471Problem D

求平均成绩Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 103770    Accepted Submission(s): 24111Problem Description假设一个班有n(n 

不容易系列之(4)——考新郎Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32532    Accepted Submission(s): 11896Problem Description国庆期间,省城HZ

海选女主角Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 52283    Accepted Submission(s): 23228Problem Descriptionpotato老师虽然很喜欢教书，但是

查找最大元素Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47246    Accepted Submission(s): 25648Problem Description对于输入的每个字符串，查找其中的最

C语言合法标识符Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 59872    Accepted Submission(s): 23694Problem Description输入一个字符串，判断其是否是C

汉字统计Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 38443    Accepted Submission(s): 20969Problem Description统计给定文本文件中汉字的个数。

统计元音Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 59563    Accepted Submission(s): 23762Problem Description统计每个元音字母在字符串中出现的次数。

Lowest Common Multiple PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 52009    Accepted Submission(s): 21589Problem Description

进制转换Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41581    Accepted Submission(s): 22802Problem Description输入一个十进制数N，将它转换成R进制数

Palindromes _easy versionTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35083    Accepted Submission(s): 21347Problem Description

人见人爱A+BTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 38164    Accepted Submission(s): 25400Problem DescriptionHDOJ上面已经有10来道A+B

杨辉三角Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 56722    Accepted Submission(s): 23663Problem Description还记得中学时候学过的杨辉三角吗？具体的

天平平衡时间限制: 1 Sec  内存限制: 128 MB提交: -  解决: -[提交][状态][讨论版]题目描述天平是一种衡器，是衡量物体质量的仪器。它依据杠杆原理制成，在杠杆的两端各有一小盘，一端放砝码，另一端放要称的物体，杠杆中央装有指针，两端平衡时，两端的质量（重量）相等。现在递归地描述一个天平：1.每个天平用两个正整数代表左右两端的质量。2.如果

暴力方法遍历求和#include<stdio.h>#include<stdlib.h>#include<time.h>#include<algorithm>using namespace std;int main() { while(1) { double j; int i,n,x=0,y=0,z=0; scanf("%d",&n);

抱着水的心态就行了=.=#include<stdio.h>#include<algorithm>using namespace std;int a[100000];int main() { int n,m,mark=1; while(scanf("%d %d",&n,&m)!=EOF) { int t,s,k=0; for(int l=0; l

暂且算简单题目吧，有点暴力-_- 用__int 64就行了#include<stdio.h>int main(){ int n,k=0; scanf("%d",&n); while(n--){ int i,j; __int64 sum=0,l; scanf("%d %d",&i,&j); for(l=i;l<

#include<stdio.h>int main() { int n,year,mouth,day; scanf("%d",&n); while(n--) { scanf("%d-%d-%d",&year,&mouth,&day); int t=0; if((year%400==0||year%4==0&&year%100!

#include<stdio.h>int main(){ int year,mouth,day,sum; while(scanf("%d/%d/%d/",&year,&mouth,&day)!=EOF) { int t; switch(mouth){ case 1:t=0; break;

辗转相除法，不多说#include<stdio.h>int mud(int p,int q) { if(p%q==0) return q; else return mud(q,p%q);}int bei(int s,int t) { return s/mud(s,t)*t;}int main() { int a,b;

a和c的最大公约数为b，所以a是b的倍数，要求最小的c，一个循环从b+1–a#include<stdio.h>int max(int a,int b) { if(a%b==0) return b; else return max(b,a%b);}int main() { int n; scanf("%d",&n); whi

博客首发，重在体验#include<stdio.h>int main() { double a,b,c,d; scanf("%lf",&d); for(int i=1;i<=d;i++) { scanf("%lf%lf%lf",&a,&b,&c); if(a+b>c&&a+c>b&&b+c>a&&a-b<c&&a-c<b&&b-c<a

入门级#include<stdio.h>int main() { double i; while(scanf("%lf",&i)!=EOF) { if(i>=0) printf("%.2f\n",i); else printf("%.2f\n",-i); } return 0;}http:/

pow函数都是浮点型参数。你也可以直接列公式#include<stdio.h>#include<math.h>#define PI 3.1415927int main() { double i,j; while(scanf("%lf",&i)!=EOF) { j=PI*pow(i,3)*4/3; printf("%.3f\n",j); }