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): 4542 Problem Description 假设有x1个字母A， x2个字母B,....

Colorful Cupcakes Time 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

Tickets Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2364    Accepted Submission(s): 1152 Problem Description Jesus, 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): 5074 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了！

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

Chopsticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1885    Accepted Submission(s): 882 Problem Description In 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): 840 Problem Description 当今国际反恐形势很严峻，特别是美国“9.11事

Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 41458    Accepted Submission(s): 17255 Problem Description Many 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): 5175 Problem 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): 4679 Problem 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) : 6 Font: Times New Roman | Verdana | Georgia Font 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*100 A2  100*5 A3 5*50 A1A2A3，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): 16478 Problem Descri