uva timecard

fromThe same family, with other method of sequencing (much more expensive than the shotgun process)That avoids the problem of repetitions. The biologists wonder if it is a waste of money the applicationOf the other method because they believe there are not a large repeated fragment in the DNA of theBacterias of the family studied.The biologists contacted you-to-write a program that, given a DNA strand, finds the largest substringThat is repeated the other times in the sequence.InputThe first li

Uva 10130-SuperSale (0-1 backpack) Question link: Click the open link Bare 0-1 backpack .. It's just like there are multiple backpacks, because these backpacks are independent of each other, and the sum is OK. #include #include #include #include #include #include #include #include #include #include #include #incl

is to find the longest common sub-sequence of two sequencesThe code is as follows:#include Uva--10066the Twin Towers +DP

Test instructions: There are N (2Analysis: From the left-hand village, whether he buys wine or sells alcohol, there is A0 traffic relative to his neighboring village, so the traffic is constantly accumulating or offsetting, counting to the right-most village.#pragma COMMENT (linker, "/stack:102400000, 102400000") #include 　　UVA-11054 Wine Trading in Gergovia (Gergovia's liquor trading) (greedy + analog)

;}DoublePolygonarea (pt* p,intN) {//Polygon Area Doubles=0; for(intI=1; i1; i++) S+ = Cross (p[i]-p[0],p[i+1]-p[0]); returns/2;}Const intN =2500+Ten; Pt P[n],ch[n];intN;intMain () {intT; scanf ("%d",T); while(t--) {scanf ("%d",N); intPc=0;Doubles1=0; Doublex,y,w,h,j; for(intI=0; i) {scanf ("%LF%LF%LF%LF%LF",x,y,w,h,K); Doubleang=-Torad (j); Pt o (x, y); P[PC++]= o + rotate (VEC (-w/2,-h/2), ANG); P[PC++]= o + rotate (VEC (w/2,-h/2), ANG); P[PC++]= o + rotate (VEC (-w/2, h/2), AN

ing A1 and A2 again. The swap map that describesthis sorting sequence is 1 2 1.For a given array, how many different swap maps exist?A Little Thoughtwill show that there is an infinite number of swaps maps, since sequentialswapping of an arbitrary pair of Elements won't change the order of the elements. Thus the Swap map 1 1 1 2 1 would also leave our arrayelements in ascending order. But what many swap maps of minimum size willplace a given array in order? That is, the question you and answer

){ - if(ioks[i][k]!=s[i-1][k]) { -ok=0; same=K; in } - if(S[i][k]>max) max=S[i][k]; to Else if(s[i][k]S[i][k]; + } - if(!ok) Break; theip[k]=s[0][k]; *netmask[k]=255; $ }Panax Notoginseng inta[Ten]={0}; - intc=max^min,sum1=0, sum2=0, sp=0; the for(intI=0;i8; i++){ + if(C>= (111))){ Asp=i+1; Break; the } + } - getbit (a,min); $ for(intI=

Longll;typedef unsignedLong Longull;Const intinf=1 -;Const DoublePi=acos (-1.0);Const intmod=1e9+7;Const intmax=200010;//twice times, because to add backwardsintMp[max],bit[max],tot,tol;//Map location Mapped tree array total sizeintLowbit (intx) { returnx (-x);}voidADD (intXinty) { while(xtol) {Bit[x]+=y; X+=lowbit (x); } return ;}intSum (intx) { intsum=0; while(x) {sum+=Bit[x]; X-=lowbit (x); } returnsum;}voidInit (intNintm) {memset (bit,0,sizeof(bit)); Tot=N; Tol=n+m; for(in

1 is "dangerous", [1,4], [1,4] and [2,2]. (? _ ?) So, should quickly change the idea! Just like sometimes the past greedy, but from the back to the greedy, instead of the first sort of r, and then L, this way to try some special data can be found.╮ (╯_╰) ╭ Well, with the code, I don't use pointers, so the code is a lot longer.1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 #defineN 50108 9 intN;Ten intVis[n]; One structnode{intl,r,id,t;} A[n],b[n]; A intSx[n],sy[n]; -

) = =2){ if(!n !m)return 0; memset (inch,0,sizeof inch); Ans.clear (); for(inti =0; i i) a[i].clear (); while(m--){ intx, y; scanf ("%d%d", x, y); A[x].push_back (y); ++inch[y]; } for(inti =1; I i) { if(inch[I] = =0) {Q.push (i); } } while(!Q.empty ()) { intt =Q.top (); Q.pop (); Ans.push_back (t); intLen =a[t].size (); for(inti =0; i i) { if(--inch[A[t][i]] = =0) {Q.push (a[t

Topic PortalTest instructions: Tell a few rectangles about the area they occupy in a convex polygonAnalysis: Training guide P272, rectangular area long * wide, as long as the calculation of all points, with convex hull and then to find the polygon area. At the center of the known rectangle, the vector rotates at the origin reference point and the angle is converted to radians./************************************************* author:running_time* Created time:2015/11/10 Tuesday 10:34:43* Fi Le N

of a[k+1]-p to make up the sum[k]+p. So from 1~sum[k+1] All can be gathered out.The implementation is the input when the sum, if it is an odd number of no solution, or else order, from large to small sweep over, select and for the SUM/2 of the symbol for the +, the rest is-.1#include 2#include 3#include 4#include 5 using namespacestd;6 7 Const intn=100010;8 structnode{intX,id;} A[n];9 intB[n],ans[n];Ten One BOOLCMP (node X,node y) {returnX.x>y.x;} A intMain () - { - intN; the Long Long

) {returnXx:y;} - intMmax (intXintY) {returnX>y?x:y;} the BOOLCMP (node X,node y) - { - if(X.Y!=Y.Y)returnx.yy.y; - returnx.xy.x; + } - intMain () + { A intT; atscanf"%d",T); - while(t--) - { -scanf"%d%d%d",n,r,l); -m=1; b[1]=0; - for(intI=1; i) in { -scanf"%d%d", a[i].y,a[i].x);// tob[++m]=a[i].x; + } -B[++m]=l;//No +1 ... theSort (A +1, A +1+n,cmp); *Sort (b +1, B +1+m); $ intp=1;Panax Notoginseng for(intI=2; i) - if(b[i]!=b[i-1]) b[

Test instructionsThere are two identical boxes, a box with n balls, another box with 1 balls, each emptying the ball less box, and then take the empty box from the other box so that after the operation of the two boxes at least one ball, judge whether it is the winner or the latter;Ideas:A SG function that takes at least one and no more than half the same at a time;AC Code:#include 　　UVA-12293 (combo game)

Test instructionsGive n heap of stones, each pile of AI, each time can take any number of stones in each heap; But the last one did not take the whole bunch away, then the next person will be in that heap to take;Ideas:Every time I get this heap, we have one strategy left.AC Code:#include 　　UVA-11892 (combo game)

Problem Solving Ideas:1. Record the molecular weight with a double array2. Store the string in a character array, scan from the back, always remember "letter", the next character to determine whether it is a number, if it is a number: With an integer record, this code with the sum, subscript + +.Judge to see if the number is recorded, that is, whether the sum has entered the while loop and is assigned, if not assigned, the next character is not a number, and the W (total record) value is directl

should all is in lower case and sorted in alphabetical order. You can is sure that he number of distinct words in the text does not exceed 5000.Sample InputAdventures in DisneylandBlondes were going to Disneyland, they came to a fork in theRoad. The sign read: "Disneyland left."So they went home.Sample OutputAAdventuresBlondesCameDisneylandForkGoingHomeInchLeftReadRoadSignSoTheTheyToBothWentWereWhenParsing: After combining the words, sort the weight.1#include string>2#include 3#include 4#includ

: Determinant one row (column) each element is multiplied by the same number and added to the corresponding element of another row (column), the determinant is unchanged.can transform the original determinant into the upper triangle or the lower triangular determinant, then the determinant value is the product of the diagonal. 1#include 2#include 3#include 4 using namespacestd;5 intN;6 inta[ *][ *];7 voiddeterminant ()8 {9 intCNT =0;Ten intAns =1; One for(inti =1; I ) A { -

", do not need to really save the tree#include#includeusing namespacestd;Const intMAXN = $;intSUM[MAXN];//Enter and count a subtrees tree with a root horizontal position of PvoidBuildintp) {intv; CIN>>v; if(v = =-1)return;//Empty TreeSUM[P] + =v; Build (P-1); Build (P+1);}//Edge read in Edge statisticsBOOLinit () {intv; CIN>>v; if(v = =-1)return false; memset (SUM,0,sizeof(sum)); intpos = maxn/2;//the horizontal position of the rootSum[pos] =v; Build (POS-1);//left dial hand treeBuild (pos +1);/

Test instructions: Given n points, your task is to make them all connected. You can create some new side, the cost equals two points distance squared (of course the smaller the better), there are several "package", you can buy, you buy, then some side can connect up,Each "package" is also spent, which allows you to find the minimum cost.Analysis: The first thought is to consider all the circumstances of the calculation, and then find the least, first Count no "package", and then some, with binar