1 solutions
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C++ :
//可参考一下java的代码,比较简洁 #include <iostream> #include <cstdio> using namespace std; int N; string A(int n) { string s=""; char t[4]; int i; for(i=1;i<=n;i++) { if(i>1) { if(i&1) s+="+"; else s+="-"; } s+="sin("; sprintf(t,"%d",i); s+=t; } for(i=1;i<=n;i++) s+=")"; return s; } string S(int n) { char t[4]; sprintf(t,"%d",N+1-n); if(n==1) return A(n)+"+"+t; else return "("+S(n-1)+")"+A(n)+"+"+t; } int main() { cin>>N; cout<<S(N); return 0; }
- 1
Information
- ID
- 17753
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
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