error 1084

A sequence of appearances refers to a series of integers with the following characteristics:D, D1, d111, d113, d11231, d112213111, ...It never equals the number D of 1, and the n+1 item of the sequence is a description of the nth item. For example, the 2nd means that the 1th item has 1 d, so it is D1;The 2nd item is 1 d (corresponds to D1) and a 1 (corresponding to 11), so the 3rd item is d111. Another example is the 4th item is d113, its description is a d,2 1, a 3,So the next item is d11231. O

1084: [usaco 3.1.6] stamps Time Limit: 1 sec memory limit: 64 MBSubmit: 122 solved: 33 submitstatusweb Board Description It is known that a set of face values (for example, {1 point, 3 points}) of N stamps and an upper limit k indicate that K stamps can be attached to an envelope. Calculate the maximum continuous postage from 1 to M. For example, suppose there are stamps of one or three points; you can paste up to five stamps. It is easy to post

); + #endif Ascanf"%d%d%d",row,col,k); the intplus=0; + for(intI=1; i) - for(intj=1; j){ $scanf"%d",a[i][j]); $sum[i][j]=sum[i-1][j]+A[i][j]; - if(a[i][j]>=0){ -plus++; thetemp+=A[i][j]; - }Wuyi } the for(intr1=1; r1) - for(intR2=1; r2) Wu for(intp=1;p ){ -F[r1][r2][p]=max (f[r1][r2-1][p],f[r1-1][r2][p]); About for(intNewr=1; Newr){ $ if(newrR1) -F[r1][r2][p]=max (f[r1][r2][p],f[newr-1][r2][p-1]+sum[r1][1]-sum

[i+1][j][2] = max (d[i+1][j][2],d[i][j][3]+a[i+1][2]);Wuyid[i+1][j+1][2] = max (d[i+1][j+1][2],d[i][j][0]+a[i+1][2]); thed[i+1][j+1][2] = max (d[i+1][j+1][2],d[i][j][1]+a[i+1][2]); -d[i+1][j+1][2] = max (d[i+1][j+1][2],d[i][j][4]+a[i+1][2]); Wu -d[i+1][j][3] = max (d[i+1][j][3],d[i][j][3]+a[i+1][2]+a[i+1][1]); Aboutd[i+1][j+1][3] = max (d[i+1][j+1][3],d[i][j][1]+a[i+1][2]+a[i+1][1]); $d[i+1][j+1][3] = max (d[i+1][j+1][3],d[i][j][2]+a[i+1][2]+a[i+1][1]); -d[i+1][j+2][3] = max (d[i

1084: [SCOI2005] Maximum sub-matrixDescriptionHere is a n*m matrix, please select the K sub-matrix, so that the K sub-matrix score of the sum of the largest. Note: The selected K-sub-matricesCannot overlap each other.Input　The first behavior is n,m,k (1≤n≤100,1≤m≤2,1≤k≤10), and the next n lines describe the score of each element in each line of the matrix (theThe absolute value of the score does not exceed 32767).Output　There is only one behavior k th

Poj 1084 Square Destroyer dlx solution repeated overwrite, pojdlx Analysis: DancingLink resolves the problem by converting it into a duplicate overwriting problem. Code: //poj 1084//sep9#include

inttemp =P.index; + while(Temp2*k) -Temp + +; the * intnum = Temp-P.index; $ if(Num >=3!Vis[temp])Panax Notoginseng { -Q.index =temp; theq.x = p.x +1; + Q.push (Q); AVis[temp] =1; the } + if(Num >=4!vis[temp-1]) - { $Q.index = temp-1; $q.x = p.x +1; - Q.push (Q); -vis[temp-1] =1; the } - if(Num >=5!vis[temp-2])Wuyi { theQ.index = temp-2; -q.x = p.x +1; Wu Q.push (Q); -vis[temp-2] =1; About

. As input, you are given a (complete or incomplete) N * n grid made with no more than 2n (n + 1) matchsticks for a natural number n Input The input consists of T test cases. the number of test cases (t) is given in the first line of the input. each test case consists of two lines: the first line contains a natural number N, not greater than 5, which implies you are given a (complete or incomplete)N * n grid as input, and the second line begins with a nonnegative integer k, the number o

I don't know why this question is classified into the search Category =! Similar to the practice of shuita. /** Tyvj-1084 digital triangle * Mike-W * 2012-3-10 * ===========================*/# include

Topic Information1084. Broken Keyboard (a)Time limit of MSMemory Limit 65536 KBCode length limit 16000 BOn a broken keyboard, some of the keys is worn out. If you are type some sentences, the characters corresponding to those keys won't appear on screen.Now given a string of you is supposed to type, and the string is actually type out, please list those keys which Is for sure worn out.Input Specification:Each input file contains the one test case. For each case, the 1st line contains the origina

* * Topic: 1, according to the number of questions answered, determine the student's score, 2, answer the number of the same, according to the time of the answer, determine the student's score; 3, answer the same number of questions, only one person can get 95,85,75,65 scores, others can only for 90,80,70,60* /# include HDU 1084 What is Your Grade?

map to record the bad keys that have occurred#include using namespacestd;stringsrc, des;intlensrc, Lendes;mapChar,BOOL>chtable;intMain () {CIN>>src; LENSRC=src.length (); CIN>>des; Lendes=des.length (); inti =0, j =0; while(I lensrc) { if(Src[i]! =Des[j]) { if(IsDigit (Src[i])) {if(!Chtable[src[i]]) {//cout printf"%c", Src[i]); Chtable[src[i]]=true; } } Else{ if(!Chtable[toupper (Src[i])) {//cout printf"%c", ToUpper (Src[i])); Chtable[toup

1#include 2 3 using namespacestd;4 5 intdp[1000001];6 7 intMain () {8 intN;9 while(Cin >>N) {Tendp[1] =1; Onedp[2] =2; A for(inti =3; I i) { - ifI2==1) -Dp[i] = dp[i-1]; the Else -Dp[i] = dp[i-1] + dp[i/2]; -Dp[i]%=1000000000; - } +cout Endl; - } + Excerpt: I is odd, split with the previous i-1 is the same, write a few groups on their own to know, do not say more. The key is that I is even when: when the split is not included in the 1 o'clock, the split

1084 matrix fetch problem V2 base time limit: 2 second space limit: 131072 KB score: 80 Difficulty: 5-level algorithm problem collection concern a m*n matrix has a different positive integer, through this lattice, you can get the reward of the corresponding value, first from the left up to the right, then from the right down to the top left. The 1th time can only go down and to the right, the 2nd time can only go up and left. Two plays if you pass th

1084. Broken Keyboard (20) [String operation] -- PAT (Advanced Level) PractiseQuestion Information1084. Broken Keyboard (20) Time limit 200 MSThe memory limit is 65536 kB.Code length limit: 16000 BOn a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters corresponding to those keys will not appear on screen. Now given a string that you are supposed to type, and the string that you are actually type out, pleas

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1084The main topic: a total of 5 questions, n students. Make 5 full marks, 0 lanes and 50 points. The students who make 1-4-way, if the first 50% (rounding down), then get 95, 85, 75, 65, if in the latter 50% will be 90, 80, 70, 60.Key IDEAS: Structure sequencingStructure sorting, consider boundary #include 　　 HDU 1084 [What is Your Grade?] struct sort

ChannelIdea: Dp[i][j][k]:k, A to I line, B to J of the maximum value of the rowCode:#include #includeusing namespacestd;intMaxPath (vectorint> > num) { Const intm =num.size (); Const intn = num[0].size (); Const intL = m + N-1; Vectorint> > g (M +1, vectorint> (M +1)); Vectorint> > H (M +1, vectorint> (M +1)); for(intK =1; K ) { intst = k 1: K-n +1; inted = k k:m; for(inti = st; I ) for(intj = i; J ) {H[i][j]= Num[i-1][k-i] +max (max (G[i-1][j-1], G[i-1][J]), Max (G[i][j

Simple question.#include #include#include#include#include#include#include#includeusing namespacestd;Const intmaxn=100000;CharS[MAXN],T[MAXN],U[MAXN];intflag[ +];intMain () {scanf ("%s", s); scanf ("%s", T); for(intI=0; s[i];i++) if(s[i]>='a's[i]'Z') s[i]=s[i]-'a'+'A'; for(intI=0; t[i];i++) if(t[i]>='a't[i]'Z') t[i]=t[i]-'a'+'A'; memset (Flag,0,sizeofflag); intp1=0, p2=0; while(1) { if(s[p1]==T[P2]) {P1++; P2++; } Else { if(flag[s[p1]]==0) {prin

); the #endif - inti,j,k; in intx, y, z the //for (scanf ("%d", cass); cass;cass--) the //for (scanf ("%d", cas), cass=1;cass About //while (~scanf ("%s", s+1)) the while(~SCANF ("%d",N)) the { thelll=aans=0; Mem (Last,0); Mem (inch,0); Mem (d,0); +scanf"%d",m); - for(i=1; i"%d%d", aa+i,bb+i); the for(i=1; i)Bayi { thescanf"%d%d",x,y); theAdd (x, y);inch[y]++; - } - Tuopu (); theprintf"%lld\n", Aans); the } the return 0; the } - /* t

1#include 2#include 3#include 4#include 5#include 6 #definell Long Long7 8 using namespacestd;9 Const intN = 1e5+ +;Ten One intA[n],dp[n]; A - voidSolve () - { the intn,k; -scanf"%d%d",n,k); - for(inti =0; I ) - { +scanf"%d",a[i]); - } +Sort (a,a+n); AMemset (DP,0,sizeof(DP)); at for(inti = n1; I >=0; i--) - { -Dp[i] =0x3f3f3f; - intj = Upper_bound (a,a+n,a[i]+2*K)-A; - if(J-i >=3) Dp[i] = min (dp[j]+1, Dp[i]); - if(J-i >=4) Dp[i] = min (dp[j-1