04 2015 档案

Digit
摘要：题目描述我们把十进制整数依次写成一个字符串，123456789101112…请问第n位数码是多少？输入第一行是一个整数T(T≤10000)，表示样例的个数。 每行输入一个整数n(1≤n≤788888899)。输出每行输出一个样例的结果。样例输入21788888899样例输出11#include#in... 阅读全文
posted @ 2015-04-26 16:01 搁浅の记忆 阅读(380) 评论(0) 推荐(1)
HDU 1133 Buy the Ticket
摘要：Problem DescriptionThe "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the ... 阅读全文
posted @ 2015-04-24 13:00 搁浅の记忆 阅读(168) 评论(0) 推荐(0)
HDU 1003 Max Sum
摘要：Problem DescriptionGiven a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7)... 阅读全文
posted @ 2015-04-24 11:07 搁浅の记忆 阅读(123) 评论(0) 推荐(0)
HDU 1702 ACboy needs your help again!（附加优先队列）
摘要：ACboy was kidnapped!! he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.As a smart A... 阅读全文
posted @ 2015-04-23 21:18 搁浅の记忆 阅读(198) 评论(0) 推荐(0)
POJ 1611 The Suspects:并查集
摘要：DescriptionSevere acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. ... 阅读全文
posted @ 2015-04-23 18:32 搁浅の记忆 阅读(132) 评论(0) 推荐(0)
矩阵
摘要：根据条件求任意子矩阵#include#include# define N 1100int a[N][N];typedef struct data{ int x, y, da;}DATA;DATA d[30];void Start(){ int i, j; for (i = 1; i = p && d... 阅读全文
posted @ 2015-04-21 16:00 搁浅の记忆 阅读(107) 评论(0) 推荐(0)
矩阵旋转
摘要：矩阵的特殊角旋转：0 90 180 270#include#includeint main (){ int T, n, m, i, j, k; char s[100][100]; scanf("%d", &T); while (T--) { scan... 阅读全文
posted @ 2015-04-20 11:43 搁浅の记忆 阅读(156) 评论(0) 推荐(0)
素数
摘要：素数1：#include#includeint Judge(int n){ int i, k; if (n == 0 || n == 1) return 0; k = sqrt(n); for (i = 2; i # define N 1000010 int a[N] = {1, 1}; void... 阅读全文
posted @ 2015-04-18 18:32 搁浅の记忆 阅读(171) 评论(0) 推荐(0)
HDU 2544 最短路
摘要：Problem Description：在每年的校赛里，所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候，却是非常累的！所以现在他们想要寻找最短的从商店到赛场的路线，你可以帮助他们吗？Input：输入包括多组数据。每组数据第一行是两个整... 阅读全文
posted @ 2015-04-18 18:00 搁浅の记忆 阅读(121) 评论(0) 推荐(0)