csu utilities

Title Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1266Descriptionfor an integerx, rememberxThe right-to-left members are respectively the first1bit, section2bit, ... IfxThe sum of the numbers on the odd digits minus the sum of the numbers on the even digits can be Onedivisible, thenxyou can be Onedivisible. Input The first line of the input data contains an integer t (1 t one row for each set of test data, with a number of digits not exceeding - a positive integer of a bit

data, the first row contains5an integerN (1 , meaning ibid. The next line containsNa no greater than -A positive integer, which in turn describes the weight of each table tennis. Outputfor each set of data, output three integers in one line, separated by spaces, indicating N a ball of Samsung, two stars and a star of the ball each have how many. Sample Input25 10 1 2 36 7 9 9 106 10 1 2 311 12 12 13 14 8Sample Output3 0 11) 3 1HINTSourceThe Seventh annual College student Program Design Competit

Title Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1407Descriptiontwo pointsA,Bare doing uniform motion. Give theT= 0 MomentsA,Bthe coordinates, andA,Bthe speed, calculationT≥0 O'Clock the minimum value of the distance between two points. Inputthe first line of input contains an integerT(1≤T≤ $), indicating that there are altogetherTGroup test Data. for each set of test data, the first row consists of 4 integersxA,yA,vAx,vAy( -103≤xA,yA,vAx,vAy≤Ten3), indicatingT= 0 MomentsAthe coordin

the swollen level 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

Owners" of 2 lives, and a "whirlwind effect" followed by 2 life-threatening "slave-owners" and 1 1-point life "terror slaves".In this series of topic 2 we have learned how to calculate how many "horrible slave owners" will be left after the N-whirlwind effect. Now that the game has a bug, the number of slave owners on the field is no longer limited by the 7 limit. There are some 1 lives left on the field, some 2 lives, some 3 life slaves, and now ask the slave owners to go through the N-whirlwi

) { theEdge[tot].next =Head[u]; *Edge[tot].to =v; $Edge[tot].index =ID;Panax NotoginsengHead[u] = tot++; - } the + voidDfsintUintPreintID) { ACnt[u][color[u]] =1; theAns[u] = Cnt[u][color[u]] 1:0; + for(inti = Head[u]; ~i; i =Edge[i].next) { - intv =edge[i].to; $ if(v = =pre) $ Continue; - Dfs (V, u, edge[i].index); - if(Cnt[u].size () cnt[v].size ()) { the swap (Cnt[u], cnt[v]); - swap (Ans[u], ans[v]);Wuyi } the for(Auto It:cnt[v]) { -

output.Sample Input19 1 4 2 6 3 8 5 9 16 2 7 6 3 5 1Sample Output3 compared to the monotonically increasing, the longest common subsequence, this problem will combine the two, very classic example. 1#include 2#include 3#include string.h>4#include 5 using namespacestd;6 Const intmaxn=1005;7 intdp[maxn],a[maxn],b[maxn],n,m;8 intLICs ()9 {Ten intI,j,max; OneMemset (DP,0,sizeof(DP)); A for(i=1; i) - { -max=0; the for(j=1; j) - { - if(A[i]>b[j] maxDp[j]) -M

--) {LL type; CIN>>type; if(type==1)//Insert{LL num; CIN>>num; if(num>v+1) {q.push (num); } Else{v+=num; while(!Q.empty ()) {LL tmp=Q.top (); if(tmp1) {v+=tmp; Q.pop (); } Else Break; } } } Elsecout1Endl; } } return 0; } /* */ /*******************************************

Test instructions: Needs to be repaired from (0,0) point to (x, y)Where there are n and y axes parallel to the riverUnit cost of road repairing C1 repair Bridge Unit Cost C2Q What is the minimum total cost?Idea: Merge all the riversThe length of the bridge repeatedlyFind the minimum cost#include 　　CSU 1548:design Road (three points)

) {if(dd[i]>0) {sum+=Dd[i]; if(dd[i]>Bb[i]) {Flag=false; Break; } } } if(!flag) {Puts ("NO"); Continue; } inthalf=len/2; Sum=0; for(intI=0;i -;++i) {if(dd[i]0) {sum+=min (bb[i],cc[i]); } Else{sum+=min (bb[i],cc[i]); } } if(sum>=half) {Puts ("YES"); Continue; } Else{puts ("NO"); Continue; } } return 0;}/**//****************************************

) { DoubleL=low,h=High ; DoubleMid= (L+H)/2, mmid= (mid+h)/2; DoubleCmid=calc (mid), cmmid=Calc (mmid); while(Fabs (Cmmid-cmid) >=1e-Ten) { if(cmid>cmmid) L=mid; Elseh=Mmid; Mid= (l+h)/2, mmid= (mid+h)/2; Cmid=calc (mid), cmmid=Calc (mmid); } returnmin (cmmid,cmid);}intMain () {intN; while(cin>>n>>x>>y>>c1>>C2) {Sum=0.0; DoubleTMP1,TMP2; for(intI=1; ii) {cin>>tmp1>>TMP2; Sum+=TMP2; } xx=x-sum; printf ("%.2lf\n", Solve (0.0, y)); } return 0;}View Cod

Topic Link: Click to open the linkTest instructionsA direction graph for a given n points (1 is the starting point and N is the end point)Each of the following lines gives the degree of a point and the next point to which it is connected.The nth point is not out of the degreeThe picture is like this: 1->2, 1->3, 2->3First question:If there is a scheme that allows this person to enter a point and no longer reach the end of the output PRISON, otherwise the output pardonSecond question:If this pers

//Dfs Augmentation Road the {Bayi for(intI=1; i) the { the if(edge[x][i]!Vis[i]) - { -vis[i]=1; the if(lin[i]==-1|| DFS (Lin[i]))//I have no matching or matching can get the augmented path the { thelin[i]=x; the return true; - } the } the } the return false;94 } the the intMain () the {98 while(SCANF ("%d", n)! =EOF) About { -Memset (Edge,0,size

Test instructions: Give some numbers if they can be divisible by 6 output beer!! Otherwise output boom!!Idea: Save 0 to 9 with a three-dimensional arrayViolence out of every number of valuesFind the results(This is a bit of a hole in the 1 number may not exist 2 number of uncertainties)#include 　　CSU 1395:timebomb (Analog)

.#include #include#include#include#includeusing namespacestd;Const intN =100005;intX[n],a[n],b[n];intC[n],n;intLowbit (inti) { returni (-i);}voidUpdateintIdxintv) { for(inti=idx;i2*n;i+=lowbit (i)) {C[i]+=v; }}intGetsum (intidx) { intsum =0; for(inti=idx;i>=1; i-=lowbit (i)) {Sum+=C[i]; } returnsum;}intMain () {inttcase; scanf ("%d",tcase); while(tcase--) {memset (c,0,sizeof(c)); scanf ("%d",N); intCNT =1; for(intI=1; i) {scanf ("%d%d",a[i],B[i]); X[cnt++] =A[i]; X[cnt++] =B[i];

; I i) {if(A[i].val = = a[i-1].val) B[i] = b[i-1]; Else if(A[i].val = = a[i-1].val +1) B[i] = b[i-1]+1; ElseB[i] = b[i-1] +2; } for(inti =1; I i) {c[A[i].pos]=B[i];} memset (tmp,0,sizeoftmp); for(inti =0; I i) {scanf ("%d%d", q[i].l,Q[I].R); Q[i].id=i; } sort (q, q+m); intPL =1, PR =0; ll ans=0; for(inti =0; I i) {intID =q[i].id; intL =Q[I].L; intR =Q[I].R; if(PR for(inti = pr+1; I 1); Else for(inti = PR; i > R; -i) ans + = Update (C[i],-1); if(PL for(inti = pl; I 1); PR= r,

); + } - returnRes; the } * voidInit () $ {Panax Notoginsengmemset (Tree,0,sizeof(tree)); - } the intMain () + { A intcas=0; the while(~SCANF ("%d%d%d",n,m,k)) + { - init (); $ for(intI=1; i) $ { -scanf"%d",a[i]); - } theSort (A +1, a+n+1); - intx, y;Wuyi intst=1, ed; the for(intI=1; i) - { Wuscanf"%d%d",x,y); -Ed=x; About intPos1=upper_bound (A +1, A +1+N,ST)-A; $ intPos2=upper_bound (A +1, A

TuneUp Utilities, believe that friends familiar with the software should have heard, the software from Germany, is recognized in the world's best system tuning tool, known as "PC system maintenance Swiss Army knife." At present, TuneUp Utilities version 2011 has been released, this issue will lead us to try TuneUp Utilities 2011, to see the world's top system opt

Evaluation Address: https://cn.vjudge.net/problem/CSU-1632Descriptionin the string, all occurrences of at least 2 number of sub-strings of the secondInputfirst act an integer T (t represents the number of use case groups, with each group of use cases occupying a length of 100000 the stringOutputFor each set of use cases, output the number of substrings that appear at least two times in the stringSample Input3AabaabAaaaaAaAaASample Output545SolutionAns

CSU 1337Time Limit:1000MSMemory Limit:131072KB64bit IO Format:%LLD %lluDescriptionFlt theorem: When n>2, the indefinite equation AN+BN=CN has no positive integer solution. For example, A3+B3=C3 does not have a positive integer solution. To enliven the atmosphere, we might as well have a funny version: Change the equation to a3+b3=c3, so there are solutions, such as a=4, B=9, c=79 43+93=793.Enter two integers x, y, to satisfy the xInputEnter up to 10 s