L-75的专栏

#include #include int a[1005][1005];int s[1005][1005];void LCSLength(char *str1,char *str2,int len1,int len2){ int i,j; for(i=1;i<=len1;i++) { for(j=1;j<=len2;j++) { if(str1[i-1]==str2[

找单词Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6496    Accepted Submission(s): 4542Problem Description假设有x1个字母A， x2个字母B,....

Colorful CupcakesTime Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^题目描述Beaver Bindu has N cupcakes. Each cupcake has one of three possible colors. In this problem we will re

TicketsTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2364    Accepted Submission(s): 1152Problem DescriptionJesus, what a grea

命运Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14387    Accepted Submission(s): 5074Problem Description穿过幽谷意味着离大魔王lemon已经无限接近了！

Function Run FunTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3358    Accepted Submission(s): 1652Problem DescriptionWe all love

ChopsticksTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1885    Accepted Submission(s): 882Problem DescriptionIn China, people u

反恐训练营Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3564    Accepted Submission(s): 840Problem Description当今国际反恐形势很严峻，特别是美国“9.11事

Bone CollectorTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 41458    Accepted Submission(s): 17255Problem DescriptionMany years

龟兔赛跑Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13906    Accepted Submission(s): 5175Problem Description据说在很久很久以前，可怜的兔子经历了人生中最

钱币兑换问题Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7866    Accepted Submission(s): 4679Problem Description在一个国家仅有1分，2分，3分硬币，将钱N兑换

计算n条直线的交点数:计算1条直线平行（即都不平行），交点为dp[n-1][j+1*(n-1)]；计算2条直线平行，交点为dp[n-2][j+2*(n-2)];                           ......计算i条直线平行，交点为dp[n-2][j+2*(n-2)];                           ......另外还有dp[i]

1、 01背包问题有n个物品，每个物品只有一件。动归方程:(1)  二维数组解法     dp[i][j]=max{dp[i-1][j-w[i]]+v[i] , dp[i-1][j]};(2)  一维数组解法     dp[j]=max{dp[j-w[i]]+v[i] , dp[j]}附代码：HDU 2602 Bone Collector二维数组：#in

Zipper Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 23   Accepted Submission(s) : 6Font: Times New Roman | Verdana | GeorgiaFont Size

最简单的最长公共子序列问题上AC的代码//最长公共子序列#include #include int a[1005][1005];void LCSLength(char *str1,char *str2,int len1,int len2){ int i,j; for(i=1;i<=len1;i++) { for(j=1;j<=len2;j++) { if(str1

最大子段和，经典的DP问题，对于刚接触DP的我，无限次的WA。一个简单的子段和问题，要注意，sum初值要设为无穷小，设为0是过不了的，还要注意，对于左右界来说，要有初值。上代码#include int main(){ int T,n; int i,j,sum,b,a,l,r,temp; scanf("%d",&T); for(j=1;j<=T;j++) { b=0;

因为矩阵乘法满足结合律不满足交换律，所以矩阵乘法进行的顺序呢不同的相乘次数就有所不同。例如{A1，A2，A2}:A1 10*100A2  100*5A3 5*50A1A2A3，有两种加括号形式，即，（A1A2）A3和A1（A2A3）；第一种乘积数乘次数为：10*100*5+5*100*50=7500第二种乘积数乘次数为：100*5*50+10*100*50=75000

Super Jumping! Jumping! Jumping!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 36136    Accepted Submission(s): 16478Problem Descri