student grade program in c

#include 9 using namespacestd;Ten One #defineN 12000 A #defineINF 0x3f3f3f3f - - intA[n]; the - intMain () - { - intt,n,i,sum,k=1; + -scanf"%d", T); + A while(t--) at { -sum=0; -scanf"%d", n); - - for(i=0; i) - { inscanf"%d", a[i]); -sum+=A[i]; to } + -printf"Case #%d:", k++); the if(sum==1) *printf"1\n"); $ ElsePanax Notoginsengprintf"%d\n", sum/2); - } the return 0; +}Attach a code that makes it clearer to people who don't un

parameters for characters//decryption for(inti =0; I ) { intj = i +K1; if(J >= Index1) J%= Index1;//because it's a rotation, don't forget to take the remainder.STR[GROUP1[J].P]=group1[i].c; } for(inti =0; I ) { intj = i +K2; if(J >= Index2) J%=Index2; STR[GROUP2[J].P]=group2[i].c; } for(inti =0; I ) { intj = i +K3; if(J >= Index3) J%=index3; STR[GROUP3[J].P]=group3[i].c; } printf ("%s\n"

Combine String#include #include#include#include#include#include#include#includeusing namespacestd;typedef unsignedLong LongLL;#defineMet (A, B) (Memset (A,b,sizeof (a)))Const intINF = 1e9+7;Const intN =2100;Const intMOD =9973;intDp[n][n];intMain () {CharS1[n], S2[n], s[n*2]; while(SCANF ("%s%s%s", s1+1, s2+1, s+1)!=EOF) { intLen1=strlen (s1+1), Len2=strlen (s2+1), I, J; intLen = strlen (s+1); ///printf ("%d%d%d\n", Len1, Len2, Len); if(len1+len2!=Len) {printf ("no\n"); Continue; }

two cases, either just well found, one on the leftLike what2 3 4 5 7 find 3 return 2Find 6 return 5 We don't know if it's on the left or the right. The difference is small, can be considered inFind 8 to return to the last position#include //#include #include string.h>#include#include#include#includeSet>#include#include#include#include#include#defineMAXN (100000+10)#defineMAXM (100000)#defineINF 0X3FFFFFFF#defineINF 0x3f3f3f3f#defineLson l,m,rt#defineRson m+1,r,rt#defineLL Long Long#defineULL un

B-Instructions for the robotTime limit:1000MS Memory Limit:131072KB 64bit IO Format:%lld %ll U Submit Status Practice CSU 1112DescriptionThere is a robot at the origin of the axis. The robot will execute a series of instructions, and your task is to predict its location after all instructions have been executed.· Left: Move one unit· Right: Move one unit· Same as I: performs the same action as article I. Input guarantee I is a positive integer and does not exceed the number of previous execut

(Point s,point e,point a,point b) {returnDirection (e,a,s) *direction (e,b,s) 0;}//where the line intersectsDoubleCalculate (Point S,point e) {inti,j,k=0; Doublesum; Point A,b,temp; for(i=0; i//traverse all points compute intersections{a=P[i]; b=p[i+1]; if(ABS (direction (e,a,s)) EPS) {Px[k++]=A; Px[k++]=b; } Else if(Intersect (S,E,A,B))//Two line intersection{px[k++]=intersection (S,E,A,B);//Two line intersection } } if(k==0) return 0.0; Sort (px,px+K,CMP);//sort

#include #include#include#includestring>#include#include#includeusing namespacestd;Chars[ -+Ten][ -+Ten];intN;mapstring,int>m;intMain () { while(~SCANF ("%d",N)) { for(intI=0; i"%s", S[i]); M.clear (); for(intI=0; i) { intlen=strlen (S[i]); for(intj=0; j) { if(s[i][j]>='A's[i][j]'Z') s[i][j]=s[i][j]-'A'+'a'; } } //for (int i=0;i intans=0; for(intI=0; i) { for(intj=i+1; j) { if(Strlen (s[i]) = =strlen (S[j])) {

#include #include#include#includeusing namespacestd;intWhp,wat,wdef,bhp,bat,bdef;Chars; intMain () { while(~SCANF ("%c",s)) {scanf ("%d%d%d",whp,wat,wdef); scanf ("%d%d%d",bhp,bat,bdef); intt1=wat-Bdef; intt2=bat-Wdef; if(s=='W') { while(1) { if(t1>0) bhp-=T1; if(bhp0) {printf ("Warrior wins\n"); Break; } if(t2>0) whp-=T2; if(whp0) {printf ("Warrior loses\n"); Break; } if(t10) {printf ("Warrior loses\n"); Break; }

Test instructionsKnown X,kAsk K to make a positive integer y for x+y==x|yX+y==x|yThat is, X is 1 bits y is 0There are several of Y can be 1See Code#include #include#includestring>#include#include#includeusing namespacestd;Const intn=50005;intn,k;intseg[ the],v[ the];Long Longm[ the];intMain () {intI,j,r; m[0]=1; for(i=1; i -; i++) M[i]=m[i-1]*2; intT; scanf ("%d",T); while(t--) {scanf ("%d%d",n,k); Long Longans=0; K intCnt=0; memset (V,0,sizeof(v)); for(i=1; i the; i++) {Seg[i]=n%2; N/

Test instructionsEnter an article that outputs up to 10 wordsDoes not include existing words in oldThe same number of occurrences by dictionary order sizeSimple simulation, number of map recordsThere's no pit.and sort it out.Use String to compare dictionary order sizes directly#include #include#includestring>#include#include#include#includeusing namespacestd;Chars[205],m[]={""};structnode{stringA; intv;} p[150005];BOOLCMP (node A,node b) {if(A.V==B.V)returna.aC.A; returnA.v>B.V;}intMain () {intI

The country ListTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 2598 Accepted Submission (s): 615Problem Descriptionas the World Expo hosted by Shanghai are coming, CC is very honorable to BES a volunteer of such an International pageant. The foreign visitors. Although he has a strong desire to is an excellent volunteer, the lack of 中文版 makes him annoyed for a long time.Some countries ' names look so similar that he can ' t distinguish them. Su

the standard Web Servie is supported.3. Reliability of security Domainws-security, Ws-trust and the ws-secureconversation are added to the SOAP message for user authentication, data integrity authentication, data privacy and many other security factors. 4. Compatibility WCF takes full account of compatibility with legacy systems. Installing WCF does not affect legacy technologies such as ASMX and. Net Remoting. Even for WCF and ASMX, although both use SOAP, WCF-developed-based applications c

Topic Link ~~>To do the idea : after the regional game did not write the tree, only remember the thought, the game thought, but only to play the code for more than half an hour (really drunk!!) ), and then is the constant debugging code, the sad reminder is the tune for one hours also did not tune out .....problem-solving ideas: This problem as long as the thought of a node of the subtree of all the node number is greater than the number of this node (in the line segment tree) and is continuous,

; A returnLow ; the } + intMain () - { $ intT, k =1; $ -stu[0].Y =1; - for(intI=1; i +; i++) the for(intJ=i; j +; J + +) - {Wuyi if(GCD (j, i) = =1) the { -stu[k].x =i; WuSTU[K++].Y =J; - } About } $Sort (Stu, stu+K, CMP); - -scanf ("%d", t); - while(T--) A { + DoubleN; thescanf ("%LF", n); - intp = Bin_sreach (n,0, K-1); $printf ("%d/%d\n", stu[p].x, stu[p].y); the

J. Ugly ProblemTest instructions: Divide the large number into a palindrome number of no more than 50The main point: each time with no more than the large number of palindrome to reduce, the results to be carried out again, note "10" The number of pit points#include #include#include#include#include#includestring>#include#include#includeSet>#includeusing namespacestd;Const Doubleeps=1e-8;intLen;voidSubtractChar*a,Char*b,Char*c) { for(inti=len-1; i>=0; i--) {C[i]=a[i]-b[i]+'0'; if(c[i]'0') c[i

Summary of the eighth ACM College Student Program Design Competition in Henan Province There are so many slots in the province and day of the year. But first, I 'd like to mention that I was so touched that I didn't eat in the dining room. The hotel's meals are more or less delicious than the school's restaurants ~~ The competition process is recorded as follows: First of all, in the warm-up match on the fi

Is ...... Determine that the intersection of the link between P and the center of the circle is not on the arc, that is, the distance from P to P, or the shortest distance between P and the endpoint. #include CSU 1503: distance from point to arc-computing program design competition for the tenth college student in Hunan Province

Phone Number Time limit:1000ms Memory limit:65536k have questions. Dot here ^_^ Topic DescriptionWe know that if a phone number A is another phone number B's prefix, B is isn't able to be called. For a example, A are 123 while B is 12345, after pressing 123, we call a, and not able to call B. Given N phone numbers, your task is-to-find whether there exits two numbers a and b that's A is B ' s prefix. inputThe input consists of several test cases. The "a" of input in each test case contains