龙腾四海365

2013年7月29日

【ACM】hdu_zs1_1003_放大的X _201307271557

摘要：放大的X Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 207 Accepted Submission(s) : 33Font: Times New Roman | Verdana | Georgia Font Size: ← →Problem Description请你编程画一个放大的’X’。如3*3的’X’应如下所示：X X XX X5*5的’X’如下所示：X X X X X X XX XInput输入数据第一行是一个整数T，表示有T组测阅读全文

posted @ 2013-07-29 15:58 龙腾四海365 阅读(191) 评论(0) 推荐(0)

【ACM】hdu_zs1_1004_第二小整数_201307271529

摘要：第二小整数 Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 14 Accepted Submission(s) : 10Font: Times New Roman | Verdana | Georgia Font Size: ← →Problem Description求n个整数中倒数第二小的数。每一个整数都独立看成一个数，比如，有三个数分别是1，1，3，那么，第二小的数就是1。Input输入包含多组测试数据。输入的第一行是一个整数C，表示有C 阅读全文

posted @ 2013-07-29 15:56 龙腾四海365 阅读(197) 评论(0) 推荐(0)

【ACM】hdu_zs1_1001_水仙花数_201307271504

摘要：水仙花数 Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 0 Accepted Submission(s) : 0Font: Times New Roman | Verdana | Georgia Font Size: ← →Problem Description春天是鲜花的季节，水仙花就是其中最迷人的代表，数学上有个水仙花数，他是这样定义的：“水仙花数”是指一个三位数，它的各位数字的立方和等于其本身，比如：153=1^3+5^3+3^3。现在阅读全文

posted @ 2013-07-29 15:55 龙腾四海365 阅读(246) 评论(0) 推荐(0)

【ACM】nyoj_103_A+BII_201307291022

摘要： A+B Problem II时间限制：3000 ms | 内存限制：65535 KB 难度：3描述 I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.A,B must be positive.输入 The first line of the input contains an integer T(1#include #define MAX_LEN 1000int an1[MAX_LEN+100];int an2[MAX_LEN+1 阅读全文

posted @ 2013-07-29 10:48 龙腾四海365 阅读(191) 评论(0) 推荐(0)

2013年7月27日

【ACM】hdu_1004_Let the Balloon Rise

摘要： Let the Balloon RiseTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 57556Accepted Submission(s): 21037Problem DescriptionContest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time 阅读全文

posted @ 2013-07-27 11:38 龙腾四海365 阅读(124) 评论(0) 推荐(0)

【ACM】poj_3981_字符串替换_201307271019

摘要：字符串替换Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8447 Accepted: 3988 Description编写一个C程序实现将字符串中的所有"you"替换成"we"Input输入包含多行数据每行数据是一个字符串，长度不超过1000 数据以EOF结束Output对于输入的每一行，输出替换后的字符串Sample Inputyou are what you doSample Outputwe are what we do#include #include int main() 阅读全文

posted @ 2013-07-27 10:24 龙腾四海365 阅读(188) 评论(0) 推荐(0)

【ACM】poj_1000_A+B_201307271012

摘要： A+B ProblemTime Limit: 1000MS Memory Limit: 10000K Total Submissions: 296408 Accepted: 162241 DescriptionCalculate a+b InputTwo integer a,b (0int main(){int a,b;while(scanf("%d %d",&a,&b)!=EOF){printf("%d\n",a+b);}return 0;} 阅读全文

posted @ 2013-07-27 10:14 龙腾四海365 阅读(95) 评论(0) 推荐(0)

2013年7月26日

【ACM】hdu_1170_Balloon Comes!_201307261946

摘要： Balloon Comes!Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16847 Accepted Submission(s): 6150Problem DescriptionThe contest starts now! How excited it is to see balloons floating around. You, one of the best programmers in HDU, can get a very b 阅读全文

posted @ 2013-07-26 20:22 龙腾四海365 阅读(234) 评论(0) 推荐(0)

【ACM】hdu_1096_A+BVIII_201307261748

摘要： A+B for Input-Output Practice (VIII)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 73526 Accepted Submission(s): 22760Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and th 阅读全文

posted @ 2013-07-26 17:50 龙腾四海365 阅读(188) 评论(0) 推荐(0)

【ACM】hdu_1095_A+BVII_201307261740

摘要： A+B for Input-Output Practice (VII)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34147 Accepted Submission(s): 22856Problem DescriptionYour task is to Calculate a + b.InputThe input will consist of a series of pairs of integers a and b, separate 阅读全文

posted @ 2013-07-26 17:40 龙腾四海365 阅读(175) 评论(0) 推荐(0)

【ACM】hdu_1094_A+BVI_201307261731

摘要： A+B for Input-Output Practice (VI)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34609 Accepted Submission(s): 23427Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains multiple test cases, and one case one li 阅读全文

posted @ 2013-07-26 17:30 龙腾四海365 阅读(151) 评论(0) 推荐(0)

【ACM】hdu_1093_A+BV_201307261715

摘要： A+B for Input-Output Practice (V)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 36633 Accepted Submission(s): 24927Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and then 阅读全文

posted @ 2013-07-26 17:22 龙腾四海365 阅读(197) 评论(0) 推荐(0)

【ACM】hdu_1092_A+BIV_201307261630

摘要： A+B for Input-Output Practice (IV)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 50209 Accepted Submission(s): 26940Problem DescriptionYour task is to Calculate the sum of some integers.InputInput contains multiple test cases. Each test case cont 阅读全文

posted @ 2013-07-26 16:37 龙腾四海365 阅读(106) 评论(0) 推荐(0)

【ACM】NYOJ_506_洗澡_20130725

摘要：洗澡时间限制：1000 ms | 内存限制：65535 KB 难度：1描述 Mostrp是个爱干净的好少年。有一次去澡堂洗澡时发现澡堂的澡柜编号中没有出现过数字‘4’。 Mostrp 感到很好奇。可能是因为在澡堂老板眼里。数字‘4’是十分不吉利的。现在Mostrp知道澡柜的最大的编号N，你能帮他算出澡堂一共有多少澡柜吗？输入有多组数据，每行输入一个N。( 1 2 int main() 3 {int n; 4 while(scanf("%d",&n)!=EOF) 5 {int i,j,m; 6 m=n; 7 for(i=1;i0)10 {if(j%10==4)1 阅读全文

posted @ 2013-07-26 16:20 龙腾四海365 阅读(177) 评论(0) 推荐(0)

【ACM】NYOJ_486_Old Calculator_20130725

摘要： Old Calculator时间限制：1000 ms | 内存限制：65535 KB 难度：1描述 szhhck have an old calculator bought 5 years ago.he find the old machine can just calculate expressions like this :A-B、A+B、A*B、A/B、A%B.because it is too old and long time not use,the old machine maybe conclude a wrong answer sometime.Your task is to 阅读全文

posted @ 2013-07-26 16:17 龙腾四海365 阅读(177) 评论(0) 推荐(0)

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