http://blog.chinaunix.net/uid-26000296-id-3322071.html前一篇文章《为FFmpeg添加自定义滤镜》详细讲述了FFmpeg的滤镜添加步骤,并给出了代码实例。本文将以FFmpeg自带的deinterlace滤镜”yadif
– yet another ...
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其他好文 时间:
2014-05-23 23:00:03
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509
编译警告:This project was created using a version of
compiler that is not currently installed: 6.0.3 [C2000]. Another version of the
compiler will be used...
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其他好文 时间:
2014-05-23 12:16:21
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3014
Color the ballTime Limit: 9000/3000 MS
(Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s):
7941Accepted Submission(s): 4070Prob...
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其他好文 时间:
2014-05-23 11:58:12
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314
Evaluate the value of an arithmetic expression
inReverse Polish Notation.Valid operators are+,-,*,/. Each operand may be an
integer or another express...
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其他好文 时间:
2014-05-22 00:49:57
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311
【题目】
Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.
For "(()", the longest valid parentheses substring is "()", which has length = 2.
Another example is ")()())", whe...
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其他好文 时间:
2014-05-20 16:39:07
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280
【题目】
Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example,
Given input array A ...
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其他好文 时间:
2014-05-18 18:25:19
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272
题目链接:hdu 4811 Ball
题目大意:有三种颜色的球若干,每次向桌子上放一个球,保证是一条序列,每次放球的得分为当前放入序列的球的前面有多少种不同的颜色a,后面的有多少种不同的颜色b,a+b。问说给定球的数量后,最大得分为多少。
解题思路:因为放球顺序是自己定的,所以我们可以尽量早得构造一个序列,使得后面放入球的得分均保持在峰值。那么求峰值就要根据球的数量来决定。我们叫得分为...
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其他好文 时间:
2014-05-18 10:01:02
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192
题意:
Evaluate the value of an arithmetic expression in
Reverse Polish Notation.
Valid operators are +, -, *,
/. Each operand may be an integer or another expression.
So...
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其他好文 时间:
2014-05-18 08:09:15
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190
【题目】
原文:
1.8 Assume you have a method isSubstring which checks if one word is a substring of another. Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using only one...
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其他好文 时间:
2014-05-16 01:50:08
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311
http://acm.hdu.edu.cn/showproblem.php?pid=4811
推理一下,发现可以先求出后面放小球可以加分的最大值,然后前面的和为0 + 1 + 2 + ...+ max,max最大为6,因为每个球最多算左右两边
代码:
#include
#include
using namespace std;
long long a, b, c;
long long ...
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其他好文 时间:
2014-05-15 23:04:28
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387