Wunder Fund Round (div. 1 + div. 2 combined)

Now the level is really not enough, only can do water problem

A. Slime Combining

Test instructions: is to give you the N 1, two the same number can be combined into a number, for example, there are two identical V, the combined value is v+1

Idea: Direct Simulation stack

1#include <iostream>2#include <algorithm>3#include <stack>4 using namespacestd;5 intnum[10000];6 intMain ()7 {8     intN;9      while(Cin >>N) {Tenstack<int>p; OneP.push (1); A      for(intI=1; i<n;++i) { -P.push (1); -         intx, y; the          while(P.size () >=2){ -x=p.top ();p. Pop (); -y=p.top ();p. Pop (); -             if(x==y) +P.push (x+1); -             Else{ + P.push (y);p. push (x); A                  Break; at             } -         } -     } -         intk=0; -          while(!P.empty ()) { -num[k++]=p.top (); in P.pop (); -         } to          for(inti=k-1; i>=0;--i) +             if(i==k-1) cout <<Num[i]; -             Elsecout <<" "<<Num[i];  thecout <<Endl; *     } $}

B. Guess the permutation

This question also thought for a long time, actually figured out is very simple

Test instructions: give you a square array of values Map[i][j]=min (A[i],a[j]), that is, the value is the smaller of the two, and array A is a sequence that allows you to restore the array A through this array of values

Idea: In each row if the number of different elements equals n, then this sequence except 0 think the other value is the number of series in the array A, the remaining as long as the 0 is changed to N.

Another way of thinking (being taught by others) is that the maximum value in each row is actually the value of the array a[i], but you need to pay attention to the n-1 N, (think)

　　

1#include <iostream>2#include <cstdio>3#include <cmath>4#include <cstring>5#include <Set>6 using namespacestd;7 Const intqq= -;8 intMAP[QQ][QQ];9 intMain ()Ten {     One     Set<int>P[QQ]; A     intN;SCANF ("%d",&n); -      for(intJ,i=1; i<=n;++i) { -          for(j=1; j<=n;++j) { thescanf"%d",&map[i][j]); - P[i].insert (Map[i][j]); -         } -     } +     intans; -      for(intI=1; i<=n;++i) +         if(p[i].size () = =N) { AAns=i; Break; at         } -      for(intI=1; i<=n;++i) { -         if(i==1) -             if(map[ans][i]!=0) printf ("%d", Map[ans][i]); -             Elseprintf"%d", n); -         Else in             if(map[ans][i]!=0) printf ("%d", Map[ans][i]); -             Elseprintf"%d", n); to     } +printf"\ n"); - } the     

#include<iostream>using namespacestd;intMain () {intn,f=1; CIN>>N;  for(intI=0; i<n;i++){        intm=0;  for(intj=0; j<n;j++)    {    intx; CIN>>x; M=Max (m,x); }    if(m==n-1&&f) {f=0; M+=1; } cout<<m<<" "; }        return 0;}

　　　　

Wunder Fund Round (div. 1 + div. 2 combined)