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DescriptionBessie noted that although humans has many universities they can attend, cows has none. To remedy this problem, she and her fellow cows formed a new university called the University of Wisconsin-farmside, "Moo U "For short."Wishing to admit dumber-than-average cows, the founders created an incredibly precise admission exam called the Cow Sc Holastic Aptitude Test (CSAT) that yields scores in the range 1..2,000,000,000.

Moo University-financial aid Question: A student in a private school applies for a scholarship, and the amount of the school is limited. Therefore, the school wants to select n logarithm from C if the amount does not exceed f. Three numbers N, C, and F are given. Returns the N logarithm in the C logarithm. Each logarithm represents the score and the applied amount respectively. If the total amount in the N logarithm is not greater than F, the interme

Moo university-financial AidTest instructionsA private school student has to apply for a scholarship, but the amount is limited. Therefore, the school would like to select N logarithm from c if the amount does not exceed F.Give three number n,c,f. Represents the n logarithm to be obtained in the C logarithm respectively.Each logarithm represents the score, with the amount of the application. Requires that the total amount of n logarithm not exceed F i

First Direct O (n²) violence. The result is a ... How weak is the usaco data = =First sort, and then you can think of a few yy ... specifically see Code-----------------------------------------------------------------------------------------#include #include #include #include #define REP (i, n) for (int i = 0; i #define CLR (x, C) memset (x, C, sizeof (x))using namespace std;const INT MAXN = 10000 + 5;int h[MAXN];int main () {//freopen ("test.in", "R", stdin);int n;cin >> N;Rep (i, N)scanf ("%d"

Title: Joseph Ring, there is a ring first delete the first element, and then every m number of deleted, ask the last left is the number 2nd element,The minimum number of M to be gone.Analysis: Number theory, simulation. The element number is 0~n-1, and the last numbered recurrence relationship is left: f (n,m) = (f (n-1,m) +m)%n.Therefore, the problem is converted into n-1 element, the first element of the Joseph Ring, in order to enumerate m to find the first one can be established.Description:

Enumeration median + priority Queue Pre-preprocessing X number of n minimum and#include #include#include#include#include#include#includeusing namespaceStd;typedefLong LongLL;Const intmaxn=100000+Ten;intN,c; LL F;structx{LL S; LL v;} T[MAXN];p riority_queueP; LL DP1[MAXN],DP2[MAXN];BOOLcmpConstXa,Constxb) { returna.sB.S;}intMain () {scanf ("%d%d%lld",n,c,f); for(intI=1; i"%lld%lld",t[i].s,t[i].v); Sort (T+1, t+1+c,cmp); LL ans=-1; if(c>=N) {if(n==1) { for(intI=1; iif(t[i].v

rectangle, and then sweep through the time of N.#include #includeusing namespacestd;Const intn=205;CharA[n][n];intN,m,i,j,k,x,y,ans,b[n][n];intMain () {scanf ("%d%d",n,m); for(i=1; i) scanf ("%s", a[i]+1); for(j=1; j) for(i=1; i) if(a[i][j]=='X') b[i][j]=b[i-1][j]+1;Elseb[i][j]=b[i-1][j]; for(i=1; i) for(j=i;j) {x=0; y=0; for(k=1; k) if(b[j][k]-b[i-1][k]==0) {x=Max (x,k); Y=x; while(x1]=='.'a[j][x+1]=='.') {x++; if(b[j][x]-b[i-1][x]==0) y=x; } ans=max (ans, (j-i+1)

0,17,16,20,19. There is breeds of cows; The first Moos at a volume of 5, and the other at a volume of 7.There is 2 cows of breed #1 and 1 cow of breed #2 in Fiel D 2, and there isAnother cow of breed #1 in field 4."Analysis" by test instructions know, each area of the sound size in addition to itself only by the previous region, minus the impact of the previous area, only the sound of their own production, the minimum number of cattle, is a complete backpack.#include #include#include#include#in

Brain-hole problems. Open the number group to record the left and right portions of each point.As long as the first point and the last point are known, the situation of the remaining points can be deduced from the efficiency of O (n).#include #includestring.h>#include#includeusing namespacestd;Const intmaxn=10000;Long LongL[MAXN],R[MAXN],A[MAXN];intMain () {intN,i; while(~SCANF ("%d",N)) {memset (L,0,sizeof(L)); memset (R,0,sizeof(R)); for(i=0; i"%lld",A[i]); Sort (A,a+N); for(I=n; i>=1; i--)

Question link: http://poj.org/problem? Id = 2231 Ideas: Sort and then derive the formula. Code: # Include Poj_2232 Moo volume

Test instructionsThere is a C with cattle, from which to choose (n-1)/2 head, so that they score the median of the largest and need financial assistance and not more than F.Analysis:The use of heaps of large top stacks maintains minimal and.Code:POJ 2010//sep9#include POJ Moo university-financial Aid Large top heap maintenance min and

of a weight. The array is scanned from left to right, And the array is scanned from right to left. A sum array is recorded. sum [I] indicates that when I is used as the median of a weight, the minimum values of the B weight and the B weight (not included in the user ). Finally, I scanned it from the back to the back and enumerated the number of the I-th ox as the median of the weight, when sum [I] is added with its own a weight less than or equal to F, the answer is output. Code: #include

Content Introduction:This prestigious C + + Classic tutorial, eight years after the end of a major upgrade. In addition to the rich practical experience of--c++ master Stanley B. Lippman, who has benefited countless programmers from around the world, Josée Lajoie's in-depth understanding of the C + + standards, and C + + pioneer Barbara E.moo in C + + Teaching insight, but also based on the new C++11 standards for a comprehensive and thorough content update. It is very commendable that all the e

// Mathematical formula derivation // sort first, it is easy to deduce and write // A1 A2 A3 A4 A5 // 1 2 3 4 5 // D1 D2 D3 D4 D5 (meaning very apparently) // 0 1 2 3 4 // 1 0 1 2 3 // 2 1 0 1 2 /3 2 1 0 1 // 4 3 2 1 0 // It is a symmetric

Test instructions A given n number. Ask what the sum of the difference between any two numbers is. Ideas: Order from small to large first. Then the DP transfer equation: Dp[i]=dp[i-1] + (i-1) * (A[i]-a[i-1]); The final result is ans=dp[n]*2;

There are C classmates, each student has the score and the Financial Aid 2 attribute, now wants you to elect the n person, satisfies their financial aid sum is less than equals F and the median number is as large as possible. I used 2 priority

"Example 7-3 text edit box to make a simple questionnaire"Geneva Ten The Run result 7-4 as shown.When you enter content in the edit box, the page changes, such as the text in the two edit boxes of the name and phone above the page disappears automatically, a "delete" icon appears on the right side of the edit box to fill in the phone information, and the contents of the edit box are automatically deleted. In addition, the contents of the two edit boxes below the page automatically increase the h

, delivery agent on a single truck, on a single delivery. The receiving clerk uses the PDA to scan the acceptance of the supplier order receipt. Every commodity from the supplier delivery, to the warehouse distribution, and then to the store receipts are traceable. Each car sent the line, the number of loading, distribution store customers are clearly clear.Euro-Hong YuanBlog: http://www.cnblogs.com/pdasoftqq:449188437Tel: 13530893964Date: September 4, 2014PDA