cisco 1260

1260: [CQOI2007] Paint paint time limit:30 Sec Memory limit:64 MBsubmit:893 solved:540[Submit] [Status] [Discuss] Description assume that you have a 5-length block that has not been painted in any color at the beginning. You want to apply its 5 unit lengths to red, green, blue, green, and red, using a string of length 5 to indicate this target: RGBGR. Each time you can paint a continuous piece of wood into a given color, the color of the coat

Time: 2016-03-27-22:37:37 Sunday Title number: [2016-03-27][hdu][1260][tickets] Problems encountered: Minute number, divided by 60 after the model T/60% 60 #include #include using namespace std; typedef long long LL; const int maxk = 2000 + 10; int s[maxk],d[maxk]; int dp[maxk]; int main(){ int t; scanf("%d",t); while(t--){ int k; scanf("%

Errorcode [1260]; % dline (s) werecutbyGROUP_CONCAT (); GROUP_CONCAT has a maximum length limit. if the maximum length is exceeded, it will be truncated. you can obtain it through the following statement: Mysql> SELECT @ global. group_concat_max_len; + ------------------------------- + | @ Global. group_concat_max_len | + ------------------------------- + | 1, 1024 | + ------------------------------- + 1024 is the default maximum length of th

Some experiences of the Huawei 1260 network adapter in ubuntu10.10-general Linux technology-Linux technology and application information. The following is a detailed description. This type of NIC is only for reference when you prefer open-source systems. Configuration: Shenzhou notebook A550-i3 cpu i3 2.13G memory 1Gx2 DDR3 1066 M hard drive west 320G 5400 to 8 M cache motherboard P55 graphics card GT240M 1024 M System: ubuntu10.10 (G desktop and K de

person, the status ② can be introduced status ①, state ①④ can be launched status ②, the status ①③ can be launched ③,①④ can launch ②.Thus the state-state transition equation:DP[1][I]=DP[2][I-1];DP[2][I]=DP[2][I-1]+DP[4][I-1];DP[3][I]=DP[1][I-1]+DP[3][I-1];DP[4][I]=DP[1][I-1];Code#include #include#include#includeusing namespacestd;intMain () {intdp[5][ -]; Memset (DP,0,sizeof(DP)); dp[1][1]=1; dp[2][2]=1; for(intI=3; i -; i++) {dp[1][i]=dp[2][i-1]; dp[2][i]=dp[2][i-1]+dp[4][i-1]; dp[3][i]=dp[1][

1, POJ 12602, Link: http://poj.org/problem?id=12603, Summary: Do not understand DP, read the http://www.cnblogs.com/lyy289065406/archive/2011/07/31/2122652.htmlTest instructions: Pearls, give demand, unit price, require the minimum amount of money can buy the same quantity, the same (or higher) quality of pearls.grasp the test instructions, 1, the input, the pearl price after input must be more expensive than the previous input. 2, with high quality pearls instead of low quality. #include #incl

interval f[i][j] to indicate to properly apply i--j this interval minimum need a few strokes, and then enumerate a k, the interval is divided into f[i][k],f[[k+1][j] two parts, F[i][j]=min (f[i][j],f[i][k]+f[k+ 1][J].But here is a special case, because each lattice can be repeatedly painted, so there will be a special sentence, with A[i] to store the color of the I lattice, if a[i]=a[j], we can initially choose to use a pen to paint them, so there is a special transfer equation: F[i][j]=min (f[

needed to buy everything on the list. Pearls can is bought in the requested,or in a higher quality class, and not in a lower one.InputThe first line of the input contains the number of test cases. Each test case is starts with a line containing the number of categories C (1The second number is the price per Pearl pi in that class (1 OutputFor each test case a single line containing a single number:the lowest possible price needed to buy everything on the LIS T.Sample Input22100 1100 231 101 111

Interval DP.#include #include#include#include#defineMAXN 55#defineINF 2000000000using namespacestd;intN,DP[MAXN][MAXN];CharS[MAXN];intMain () {scanf ("%s", s); n=strlen (s); for(intI=1; i) for(intj=1; j) { if(I==J) dp[i][j]=1; Else if(Iinf; Elsedp[i][j]=0; } for(intL=2; l) for(intI=1; i1; i++) { intleft=i,right=i+l-1; for(intk=left;k1; k++) Dp[left][right]=min (dp[left][right],dp[left][k]+dp[k+1][right]-(s[left-1]==s[right-1])); } printf

A nudnik photographer Note: Sort the numbers from 1 to n. 1 must be at the leftmost. The difference between two adjacent numbers cannot exceed 2. How many sort methods are available. Ideas: DP for the seat, when the first is 1, the second is 2, the combination of DP [I-1]; when the first is 1, the second is 3, the third is also determined to be 2, combined as DP [I-3]; and the last case is 1357 ...... 8642. So the DP equation is DP [I] = DP [I-1] + dp [I-3] + 1. 1 # include URL

Pearls Time limit:1000MS Memory Limit:10000KB 64bit IO Format:%i64d %i64 U Submit Status Practice POJ 1260 DescriptionIn Pearlania everybody is fond of pearls. One company, called the Royal Pearl, produces a lot of jewelry with pearls in it. The royal Pearl have its name because it delivers to the royal family of Pearlania. But it also produces bracelets and necklaces for ordinary people. Of course the quality of the pearls for these p

[Description ]: Input: 2220 254018 Output: The data here is represented in sequence as follows: the first 2 represents two groups of data, and the second represents two people. If you buy a ticket, the first person will spend 20 s, the other person will spend 25 s. If the two buy together, it will take 40 s (note that the two buy together must be two matching talents ), because the question is how long is the shortest time, enter 40 s. The specific time is: 08:00:40 amThe other one will not g

HDU 1260: Tickets (DP), hdu1260ticketsdp Tickets Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission (s): 923 Accepted Submission (s): 467 Problem DescriptionJesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when cocould he go back home as early as possible. A good approach, cing the total time o

Tags: bzoj bzoj1260 Dynamic PlanningGiven a piece of wood board, each of the above positions has a color. Ask how many times the brush can reach this color sequence.For dynamic planning, you can re-process it first (not necessary), so that f [I] [J] indicates the minimum number of times that J positions starting with I are flushed into the corresponding color sequence, the status transfer is as follows:If s [I] = s [J] Then f [I] [J] = min (F [I-1] [J], F [I] [J-1]) combine I and the right half

POJ 1260-Pearls (DP)Pearls Time Limit:1000 MS Memory Limit:10000 K Total Submissions:7465 Accepted:3695 DescriptionIn Pearlania everybody is fond of pearls. one company, called The Royal Pearl, produces a lot of jewelry with pearls in it. the Royal Pearl has its name because it delivers to the royal family of Pearlania. but it also produces bracelets and necklaces for ordinary people. of course the quality of th

(1Outputfor every scenario, please tell Joe at-time could he go back home as early as possible. Every day Joe started he work at 08:00:00 am. The format of time is HH:MM:SS am|pm.Sample Input2220 254018Sample output08:00:40 am08:00:08 amSOURCE Zhejiang University of Technology fourth session of college students Program design contestrecommendjgshining | We have carefully selected several similar problems for you:1257 1160 1231 1074 1069#include #include#include#include#include#includeusing name

/******************************2 code by drizzle3 blog:www.cnblogs.com/hsd-/4 ^ ^ ^ ^5 o o6 ******************************/7 //#include 8#include 9#include Ten#include One#include A#include -#include - #definell Long Long the #definePI ACOs (-1.0) - #defineMoD 1000000007 - using namespacestd; - intT; + intN; - intcost[ the]; + intqu[ the]; A intsum[ the]; at intdp[ the]; - intMain () - { -scanf"%d",t); - for(intI=1; i) - { inscanf"%d",n); -sum[0]=0; to for(intj=1; j) +

The formula is very simple, that is, the initial value should be handled well.#include HDU 1260 Tickets

format of time is HH:MM:SS am|pm.Sample Input2220 254018Sample output08:00:40 am08:00:08 amSOURCE Zhejiang University of Technology fourth session of college students Program design contestrecommendjgshining | We have carefully selected several similar problems for you:1257 1231 1074 1069 1159 DP State transfer equation for Dp[i]=min (dp[i-1]+a[i],d P[i-2]+b[i]) Dp[i] The minimum time it takes for a person to buy a ticket. A[i] and B[i] separately to buy tickets separately and two people to buy

State transfer: dp[i] = min (Dp[i], dp[j] + price[i]* (sum[i]-sum[j]+10))　　　　1#include 2#include 3#include 4 using namespacestd;5 Const intN =101;6 7 intN;8 intT;9 intSum[n], dp[n], price[n];Ten One intMain () { Ascanf"%d", t); - while(t--) { -scanf"%d", n); the for(inti =0; I ) { -scanf"%d%d", sum[i], price[i]); - if(I >0) { -Sum[i] + = sum[i-1]; + } - } + A for(inti =0; I ) { atDp[i] = (Sum[i] +Ten) *Price[i]; - for(intj =0;