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Search a 2D Matrix leetcode java
题目:Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:Integers in each row are sorte....
分类:编程语言   时间:2014-07-22 00:35:34    阅读次数:290
高斯消元模板
1.要求系数矩阵可逆2.A为增广矩阵,即A[i][n]为第i个方程右边的bi3.运行结束后A[i][n]即为第i个未知数的值typedef double Matrix[N][N];void gauss_elimination(Matrix A,int n){ int i,j,k,r; f...
分类:其他好文   时间:2014-07-22 00:27:38    阅读次数:187
HDOJ 1001 Sum Problem
Sum Problem Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 275913    Accepted Submission(s): 68867 Problem Description Hey, welcome...
分类:其他好文   时间:2014-07-22 00:27:36    阅读次数:167
HDOJ 1040 As Easy As A+B
As Easy As A+B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 36141    Accepted Submission(s): 15642 Problem Description These day...
分类:其他好文   时间:2014-07-22 00:26:35    阅读次数:233
HDOJ 4858 项目管理
暴力... 项目管理 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 537    Accepted Submission(s): 189 Problem Description 我们建造了一个大...
分类:其他好文   时间:2014-07-22 00:19:35    阅读次数:273
杭电2084 数塔
/***** HDOJ_2084_数塔问题 ********//******** written by C_Shit_Hu ************/ ////////////////动态规划DP问题/////////////// /*******************************.....
分类:其他好文   时间:2014-07-22 00:10:33    阅读次数:159
HDOJ 2002 计算球体积
计算球体积 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 79807    Accepted Submission(s): 32794 Problem Description 根据输入的半径值,计算球的体积。...
分类:其他好文   时间:2014-07-22 00:06:36    阅读次数:237
HDOJ——2072单词数
SSCANF用法:(继qsort,bsearch,strchr后发现的又一好使的函数)sscanf与scanf类似,都是用于输入的,只是后者以键盘(stdin)为输入源,前者以固定字符串为输入源。例子: 1. 常见用法。char buf[512] ;sscanf("123456 ", "%s", b...
分类:其他好文   时间:2014-07-21 10:04:37    阅读次数:302
HDOJ 4868 逃生
BestHack.....真乱..... 逃生 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 139    Accepted Submission(s): 20 Problem Description ...
分类:其他好文   时间:2014-07-20 23:21:53    阅读次数:276
UVA 11149 - Power of Matrix(矩阵倍增)
UVA 11149 - Power of Matrix 题目链接 题意:给定一个n*n的矩阵A和k,求∑kiAi 思路:利用倍增去搞,∑kiAi=(1+Ak/2)∑k/2iAi,不断二分即可 代码: #include #include const int N = 45; int n, k; struct mat { int v[N][N]; mat() ...
分类:其他好文   时间:2014-07-20 23:07:20    阅读次数:274
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