Codeforces Round #268 (Div. 1)

A:24 Game

Test instructions: Give an n, ask from 1 to n by subtraction to get 24 of the method.

Solution: It is not difficult to think of n<4 time must be without solutions,

N=4 Time 1+2=3 3+3=6 6*4=24

N=5 time 1*5=5 5-2=3 3+3=6 6*4=24

When n>5 we can let N-(n-1) =1 1*1=1 so 1~n into 1~n-2

So for any given n, we will change the number of (5~n) in the adjacent subtraction into 1 final conversion with the 1~4 or the structure of the answer can be.

1#include <cstdio>2 3#include <cmath>4 5#include <iostream>6 7#include <algorithm>8 9#include <cstring>Ten  One  A  - using namespacestd; -  the intN; -  - voidWorkintx) -  + { -  +     if(x==4)  A  at     { -  -printf"1 + 2 = 3\n3 + 3 = 6\n4 * 6 = 24\n"); -  -         return; -  in     } -  to     if(x==5) +  -     { the  *printf"1 * 5 = 5\n5-2 = 3\n3 + 3 = 6\n6 * 4 = 24\n"); $ Panax Notoginseng         return; -  the     } +  Aprintf"%d-%d = 1\n1 * 1 = 1\n", x,x-1); the  +Work (X-2); -  $ } $  - intMain () -  the { - Wuyiscanf"%d",&n); the  -     if(n<4) Wu  -     { About  $printf"no\n"); -  -         return 0; -  A     } +  theprintf"yes\n"); -  $ Work (n); the  the     return 0; the  the}

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B:two Sets

Test instructions: Given n number P1~PN requirements to divide this n number into two sets, the requirement if X∈a a-x∈a if X∈b is b-x for given.

Puzzle: Use a map to record whether each number has occurred, if for a pi A-pi is not in this number, we can only put Pi into the B set, so with a queue record must appear in the B set number,

Each time the team head element X is removed, check whether the b-x appears in P, if it appears to join this queue, and then join the b-x at the same time easy to know A-(b-x) must also appear in set B, but also to join the queue.

1#include <cstdio>2 3#include <iostream>4 5#include <algorithm>6 7#include <cmath>8 9#include <cstring>Ten  One#include <queue> A  -#include <map> -  the using namespacestd; -  -map<int,int>Pos; -  + intn,a,b,ans[500000],x[500000]; -  + intMain () A  at { -  -queue<int>Q; -  -scanf"%d%d%d",&n,&a,&b); -  in      for(intI=1; i<=n;++i) -  to     { +  -scanf"%d",&x[i]); the  *pos[x[i]]=i; $ Panax Notoginseng     } -  the      for(intI=1; i<=n;++i) +  A     { the  +         if(pos[a-x[i]]==0) -  $         { $  -ans[i]=1; -  the Q.push (X[i]); - Wuyi         } the  -     } Wu  -      while(!q.empty ()) About  $     { -  -         intnow=Q.front (); -  A Q.pop (); +  the         if(pos[b-now]==0)  -  $         { the  theprintf"no\n"); the  the             return 0; -  in         } the  the         if(ans[pos[b-now]]==1)Continue; About  theans[pos[b-now]]=1; the  the         if(pos[a-b+now]!=0) +  -         { the Bayians[pos[a-b+now]]=1; the  theQ.push (a-b+Now ); -  -         } the  the     } the  theprintf"yes\n"); -  the      for(intI=1; i<=n;++i) printf ("%d", Ans[i]); the  the     return 0;94  the}

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C:hack it!

Test instructions: Defines the sum of the numbers f (x) for x, such as f (1234) =1+2+3+4=10, which is calculated by one of the following algorithms

sum=

ans = Solve (l, r)% A;
if (ans <= 0)
ans + = A;

When the ans=0 error occurs, we construct a l,r to hack the given a.

Gy god Ben told me.

Let's look at the sum value of l=1 r=1e18.

L R When I was +1 we found that the sum value was added to 1! Because R plus 1 after the next few is the number of L does not add 1 times, so each add 1 is the whole sequence forward to push a bit, and then r into 1000 ... L So l,r each add 1,sum add 1, so first find out l=1 r=1e18 sum value after move (a-sum) times can! R is large enough to do this without having to consider the rounding problem, otherwise R is not 1 at the front, and there is no guarantee that L will appear behind. Cleverly constructed ~

Algorithm for sum: Sum value of ai= (1~10^i)-1

A1=45

A2=10*a1+45*10

...

An=10*an-1+45*10^i

Get an=45*n*10^ (n-1)

A18=81*1e18

So sum= (a18+1)%a; the problem is solved.

1#include <cstdio>2 3#include <cmath>4 5#include <iostream>6 7#include <cstring>8 9#include <algorithm>Ten  One  A  - using namespacestd; -  thetypedefLong LongLL; -  - LL L,r,a,ans; -  + intMain () -  + { A  atL=1; r=1e18; -  -scanf"%i64d",&a); -  -Ans= ((3*(3*(3*(3*(5*(2* (LL) 1e17%a))%a)%a)%a)%a)%a)%a+1)%A; -  inprintf"%i64d%i64d", l+a-ans,r+a-ans); -  to     return 0; +  -}

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Codeforces Round #268 (Div. 1)