#include #define _overload(_1,_2,_3,name,...) name #define _rep(i,n) _range(i,0,n) #define _range(i,a,b) for(int i=int(a);i=int(b);--i) #define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__) #define _all(arg) begin(arg),end(arg) #define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg)) #define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary) #define clr(a,b) memset((a),(b),sizeof(a)) #define bit(n) (1LL<<(n)) #define popcount(n) (__builtin_popcountll(n)) using namespace std; templatebool chmax(T &a, const T &b) { return (abool chmin(T &a, const T &b) { return (bEPS=1e-8 [-10000,10000]->EPS=1e-7 inline int sgn(const R& r){return(r > EPS)-(r < -EPS);} inline R sq(R x){return sqrt(max(x,0.0));} const int dx[8]={1,0,-1,0,1,-1,-1,1}; const int dy[8]={0,1,0,-1,1,1,-1,-1}; // Problem Specific Parameter: // Description: Chinese Remainder Theorm // TimeComplexity: $\mathcal{O}(\log|m|)$ // Verifyed: CF 724 C inline tuple extgcd(ll a,ll b){ ll x=1LL,y=0LL,g=a; if(b!=0){ tie(g,y,x)=extgcd(b,a%b); y-=a/b*x; } return make_tuple(g,x,y); } const int limit=10000; int main(void){ ll x,y,z; cin >> x >> y >> z; ll a,b,g; tie(g,a,b)=extgcd(x,y); string ans="mourennaihasimasenn"; if(z%g==0){ a*=z/g,b*=z/g; while(abs(a)+abs(b)>abs(a-y)+abs(b+x)) a-=y,b+=x; while(abs(a)+abs(b)>abs(a+y)+abs(b-x)) a+=y,b-=x; if(abs(a)+abs(b)>limit){ ans="mourennaihasimasenn"; }else if(a > 0 and b > 0){ ans+=string(abs(a),'c'); ans+=string(abs(b),'w'); ans+=string(abs(a)+abs(b)-1,'C'); }else if(a > 0 and b < 0){ ans+=string(abs(b),'w'); ans+=string(abs(a),'c'); ans+=string(abs(a)-1,'C'); ans+=string(abs(b),'W'); }else if(a < 0 and b > 0){ ans+=string(abs(a),'c'); ans+=string(abs(b),'w'); ans+=string(abs(b)-1,'C'); ans+=string(abs(a),'W'); } } if(ans.size()>limit) ans="mourennaihasimasenn"; cout << ans << endl; return 0; }