这套题很有意思2333

蠢了……首先先判总共加起来等不等于\(n\)，不是的话就不行

然后dfs记录\(n\)不断分下去能分成哪些数，用map记录一下，判断是否所有数都能被分出来就是了

//minamoto#include
    
     #define int long long#define fp(i,a,b) for(register int i=a,I=b+1;i
     
      I;--i)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    int res,f=1;char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}const int N=1e5+5;int n,sum,x,m,a[N];map
      
       mp;void dfs(int n){    if(mp[n])return;mp[n]=1;    dfs(n/2),dfs(n-n/2);}signed main(){//  freopen("testdata.in","r",stdin);    n=read(),m=read();    fp(i,1,m)a[i]=read(),sum+=a[i];    if(sum!=n)return puts("ham"),0;    dfs(n);fp(i,1,m)if(!mp[a[i]])return puts("ham"),0;    return puts("misaka"),0;}