bronze balloons

The skin fair mm is very sweet water, but the bronze skin mm can show individuality, the same attention, let this summer no longer thousand, no longer boring! This tutorial mainly uses Photoshop simple to pull out the character copper color effect, this summer, "Transformers" set off an irresistible upsurge, the skin fair mm is very sweet water, but the bronze skin mm can be personalized, the same attention

Given n Balloons, indexed from 0 to n-1 . Each balloon are painted with a number on the IT represented by array nums . You is asked to burst all the balloons. If The burst balloon you'll i get nums[left] * nums[i] * nums[right] coins. Here and is left right adjacent indices of i . After the burst, the and then left right becomes adjacent.Find The maximum coins you can collect by bursting the

Problem description Goddess invited acdream to have a gala. Obviously, as ACM's leader, there are no fewer balloons ~. Goddess has prepared three colors of balloons, red, yellow, green (Traffic lights ?) A balloon cannot satisfy the goddess. The goddess must write on the balloon. What should I do ~? The ghost of string Shenma loves ~ Goddess first took out a string, and then wrote every real prefix of the s

method: First in the middle of the fall, if broken, indicating the target bit Placed in the lower part, the words are not broken in the upper half. Then continue in the corresponding part of the next two points down ... Need Logn time. But this is a limited amount of water polo, for example, if you have a water polo, you just put it down on the middle floor and if you break it, then you don't have any other balls to do the experiment. So you can only start from the first floor to throw up, th

Portal $ N $ balloons are given, and $ X_ I $ and $ r_ I $ are given from left to right. They are now inflated from left to right. Each balloon is always tangent to the ground at the $ X_ I $ point during the inflation process, and the maximum radius is $ r_ I $. If the balloon is tangent to a previous balloon during inflation, stop inflation. Ask the radius of each balloon. $ N \ Leq 2 \ times 10 ^ 5, X_ I, r_ I \ Leq 10 ^ 9 $, to ensure that $ X_ I

Shanghai, ACM-ICPC national invitational Competition ----- Acmer Some people may wonder why the late Bronze medal was delayed by the organizer? Or is it a little late to hold the Shanghai Invitational competition? Not actually. Why is it the late Bronze Medal? The reason why I call it late actually has two meanings. First of all, this bronze medal was played out

) Map [I] [J] = 1; } Sum = 0; For (I = 1; I { Memset (mark, 0, sizeof (Mark )); If (find (I )) Sum ++; } Printf ("% d \ n", sum ); } Return 0; } You can first perform HDU 2119. You can only delete one row or one column at a time. The result is how many times can 1 be deleted, and a behavior subset is used as a subset, if map [I] [J] = 1 is to create an edge between I and j and delete 1, that is, each edge is to be deleted, that is, the minimum point coverage problem, this is also the question t

Test instructions is easy to understand: Limak can give their own balloons to others, so that other people's balloons than weight. Thus allowing others to float on the ceiling, unable to participate in the rankings, thus making limak their rankings rise. This topic is a typical greedy, when limak his own balloon to others, his own hands of the balloon will be reduced, then his own ranking is likely to be re

The industry mainstream PLM products are based on the history of PDM system extension and development, more emphasis is the product itself delivery management (BOM, drawings, models, etc.), into the PLM era, around the product itself launched the front-end planning, intermediate project management, follow-up service management become the focus of business promotion Bronze RDM in scientific research project management flexibility has more than more tha

Title DescriptionDescriptionEnter n number, nEnter a descriptionInput DescriptionFirst line an integer nNext line n integers, each integer less than 231-1Output descriptionOutput DescriptionMinimum and maximum numberSample inputSample Input41 2 3 4Sample outputSample Output1 4Data range and TipsData Size HintNoIdea: This problem is relatively simple, you can put all the numbers into the array, and then a little use of the fast line can be finished, the output of the No. 0 and the first n-1The c

outputSample Output620Data range and TipsData Size HintThe original ticket price is greater than 100 less than 9999, the discount rate is greater than 1 less than 9.9.Idea: This problem is a bit complicated, you can save the ticket price and discount rate with two float floats, then calculate the product divided by 100 (why divide by 100, because to keep to 10 bits, if divided by 10 is only reserved to the bit, and then the result is returned to 10), you can use: = (int) (x+0.5 ) > (int) x? (i

Some water problems.T1201:#include intmin=0x7fffffff, max=-0x7fffffff;intN,i,a;intMain () {scanf ("%d",n); for(i=1; i) {scanf ("%d",a);if(A>max) max=A;if(aA;} printf ("%d%d\n", Min,max);}View CodeT1202:#include intsum=0;intN,i,a;intMain () {scanf ("%d",n); for(i=1; i) {scanf ("%d",a); Sum+=A;} printf ("%d\n", sum);}View CodeT1203:#include #includeusing namespacestd;Const Doublee=1e-8;intMain () {DoubleA, b;cin>> a >>b;if((a-e"Yes";Elsecout"No";}View CodeT1206:#include double A; int Main () {sca

The Lunar New Year of July is the most romantic Chinese Traditional festival "Tanabata festival", but also the legend of the Cowherd and Weaver in Bridge meet days, it is said that people can also steal under the fruit racks to meet the whispers in heaven. The myth of Bridge is the only place the lovers meet in a year, and we use Photoshop to make a bridge bronze engraving. Making tools: Photoshop CS, other version steps similar. Production process:

Long time no title has been no state and then brush a water problem to playWinter vacation Training and school games have done a similar topic but then did not (in fact, now will notTest instructions: There are k balloons and an n-storey building, the balloon has a hardness, in some would have just broken, ask at least how many times the experiment can be found just broken floor.Solution: Divided into two cases: the current floor broken, the current f

Source: http://acm.fzu.edu.cn/problem.php? PID = 1515 Fuzhou University oj15 There is a cube box that provides the coordinates of two vertices on the diagonal corner. in this box, there are N points, where n Because there are only 6 balloons in total, there are 6 in total! = 720 expansion schemes. it is easy to think of enumeration because there are few cases. use DFS to find the order of each expansion scheme, and then obtain the distance from eac

The idea of the problem is very difficult to think about. Asking the minimum number of experiments you need is very difficult to solve. And we know that there are only three conditions. K, N, number of experiments.So we'd better change our mind and turn to the highest number of floors we can determine. Then use D[I][J] to indicate the maximum number of floors that can be determined with the I-ball and the J-time of the experiment.So if we're on the K-floor, there are two possibilities: 1, the ba

2018-10-03 19:29:43Problem Description:Problem solving:Very interesting topic, the first thought is the violent traversal solution space, of course, also used to memo, unfortunately or tle, because time complexity is indeed a bit too high, it should be O (n!). MapThe main subject gives the data scale, has already indicated that the time complexity is O (n^3) around more appropriate. The standard solution of the subject is given below, and the dynamic programming algorithm is used.Dp[i][j]:nums[i

#include #includeusing namespacestd;Const intMAXK = -;Const intMaxa = the; unsignedLong Longd[maxk+1][maxa+1];intMain () {memset (d,0,sizeof(d)); for(inti =1; I ) for(intj =1; J ) D[i][j]= d[i-1][j-1] +1+ d[i][j-1]; intK; unsignedLong LongN; while(Cin >> k >> N k) {intAns =-1; for(inti =1; I ) if(D[k][i] >=N) {ans=i; Break; } if(Ans 0) cout"More than""Trials needed.\n"; Elsecout"\ n"; } return 0;}uva10934 Dropping water balloons

, Tan2, tan=0; intI, J, Z, Lnor, Lrot, X, y; Init (); for(i=1; ii) for(j=1; jj) {Node No, RO; Lrot=lnor=MAXN; //Normal for(z=0;z8; z+=2) Lnor=min (Lnor, DP[Z][I][J].L), no=no+Dp[z][i][j]; for(z=0;z8; z+=2) { if(dp[z][i][j].l>Lnor) {x=i+dx[z]* (Lnor); y=j+dy[z]*(Lnor); No=no-Dp[z][x][y]; } } //Rotate for(z=1;z8; z+=2) Lrot=min (Lrot, DP[Z][I][J].L), ro=ro+Dp[z][i][j]; for(z=1;z8; z+=2) { if(dp[

AI and water level bi for the flood of the first I, and count how many bridges have been flooded at least k times. The initial water level is 1, and the swollen water level of each flood must be greater than the water level of the last flood.InputThe input file contains up to 25 sets of test data. The first behavior of each group of data is three integers n, m, K (1OutputFor each set of data, output the number of bridges that have been flooded at least k times.Sample Input2 2 22 56 28 35 3 22 3