general electric blender

]==true ) { while( K!=0) {Sum+=K%Ten;K/=Ten; } if( Prime[Sum]!=true) {Beauty[I]=false; } }}} voidDp()///Dynamic planning idea, dp[i] that is, the number of primes 1 to I {Memset(Dp,0 ,sizeof( Dp)); for (int I=1;I1000000;I + +) { if( Beauty[I]==true) {Dp[I]=Dp[I-1]+1; } Else Dp[I]=Dp[I-1]; }}int main() { int A,B,T,_count;Findprime();Beautyprime();Dp(); while (scanf("%d",T)!=Eof) {_count=0; while (T--) {_count++;scanf("%d%d",A,B);Printf("Case #%d:%d\n",_co

Output010SourcenoiRecommendll | We have carefully selected several similar problems for you:1404 1536 1517 1524 1729 Witzov game (Wythoff game): There are two piles of various items, two people take turns From a heap or at the same time from the two piles to take the same number of items, the provisions of at least one at a time, more than open, the last to win the light.In this case it is quite complicated. We use (AK,BK) (AK≤BK, k=0,1,2,...,n) to indicate the number of items and call it the s

Problem number Hdu Type Game Distribution Eventually 1001 5327 YES √ 1002 5328 YES √ 1003 5329 H 1004 5330 W 1005 5331 H 1006 5332 W 1007 5333 H 1008 5334

The meaning of the title is: the Rabbit casually hide in a hole, the wolf from the hole No. 0 began to find, every time to find the next is a number of M, all the holes formed ring.Input m and n, Hole is number No. 0 to n-1.The greatest common divisor of M and N is 1, and the wolf can be found in every hole. Otherwise, some holes are ignored, and those holes are safe.The following is the code for the AC:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master per

hook after performing the operations.Consider the original hook is a made up of cupreous sticks.Inputthe input consists of several test cases. The first line of the input is the number of the cases. There is no more than cases.For each case, the first line contains a integer N, 1Next Q lines, each line contains three integers X, Y, 1Outputfor, print a number in a line representing the total value of the hook after the operations. Use the format in the example.Sample Input11021 5 25) 9 3Sample O

Problem description for a given string, statistics the number of occurrences of a numeric character.Input data has multiple lines, the first line is an integer n, indicating the number of test instances, followed by n rows, each containing a string consisting of letters and numbers.Output for each test instance, outputs the number of values in the string, with each output occupying one row.Sample INPUT2ASDFASDF123123ASDFASDFASDF111111111ASDFASDFASDFSample Output69Code:1#include 2#include malloc.

Nuclear reactorTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 12219 Accepted Submission (s): 5547Problem description There are two types of events occurring in a nuclear reactor:When the high-energy particle strikes the nucleus, the particle is absorbed, releasing 3 high-energy particles and a low-energy particle.When the low-energy particle strikes the nucleus, the particle is absorbed, releasing 2 high-energy particles and a low-energy par

+ + and use __int64 against VC6AuthorUnknownSource 2006/1/15ACM Program Design Final ExamRecommendLcyAnalysis: From the No. 0 second to 1 seconds, 1 high-energy particles emit 3 high-energy particles and a low-energy particle, from the 1th second to 2nd seconds 3 high-energy particles and a low-energy particle generation (3*3+2*1) a high-energy particle, while 3 high-energy particles produce 3 low-energy particles + 1 low-energy points to produce 1*1; a[ The array contains a high-energy particl

Find the safest roadTime limit:10000/5000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 9368 Accepted Submission (s): 3308Problem Descriptionxx Planet There are many cities, each city has one or more flight lanes, but not all roads are very safe, each road has a safety factor s,s is the real number between 0 and 1 (including 0, 1), a channel p from u to V The safety is safe (P) = S (E1) *s (E2) ... *s (EK) E1,e2,ek is on the side of P, now 8600 want to go out t

A = = B?Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 64239 Accepted Submission (s): 10060Problem descriptiongive You both numbers A and B, if A is equal to B, you should print "YES", or print "NO".Inputeach test Case contains-numbers A and B.Outputfor each case, if A was equal to B, you should print "YES", or print "NO".Sample Input1 22 23) 34 3Sample OutputNoyesyesno/* The first time to do is always to understand the topic too simple. Aft

*root +1, Mid +1, right); at returnTree[root].max =Max (A, b); - } - intUpdate (intRootintPosintval) - { - if(Pos //the point is not in a child segment - returnTree[root].max; in if(pos = = Tree[root].left pos = = tree[root].right)//leaves; - returnTree[root].max =Val; to intA, B; +A = Update (2*Root, Pos, Val); -b = Update (2*root+1, POS, Val); theTree[root].max =Max (A, b); * returnTree[root].max; $ } Panax Notoginseng intFind (intRootintLeftintRight ) -

The meaning of the topic: to the point interval [a, b], find the number of K, and POJ2104, just the time limit on the Hdu 5000MS, with my method on the POJ, too, will time out.The code of this problem, change the input of the main function, you can directly AC the POJ on the 2104.This problem, with the division of the method, WR, Tangled up one night, finally gave up, really do not know where the wrong. Then the method of dividing tree was used to learn the establishment and search of the partit

This problem, similar to the decryption of ciphertext. The meaning of the topic is very clear.Give you the number of P Q e l,l for the second row.Calculate the N and FN first, and then calculate the D.n = p * q;fn = (p-1) * (q-1).Then according to the D * e% fn = 1, calculate d.The key is to seek C. The number num to you is equal to c ^ d% n. c The corresponding ASCLL code is the decrypted character.The following is the code for the AC:#include Hangzhou Elec