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听学长说了这场比赛十分有意思，所以来玩玩。 学长真的强，啥都会，全程听正解然后码代码.....
A.Numbers Joke
只有一个样例，输入3输出27 机智的学长瞬间就在OEIS上找到了
#include<iostream> #include<cstdio> #include<cmath> #include<queue> #include<cstring> #include<queue> #include<algorithm> #include<map> #define ll long long using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f; } int s[]={4,22,27,58,85,94,121,166,202,265,274,319,346,355,378,382,391,438,454,483,517,526,535,562,576,588,627,634,636,645,648,654,663,666,690,706,728,729,762,778}; int main() { int n=read(); cout<<s[n-1]; return 0; }
B.Kids' Riddle
只给输入输出 input 11 output 2 input14output0 input61441output2 input571576output10 input2128506output3
我也不知道学长是怎么发现的，正解是转换成16进制，然后数圈圈（对，你没看错）
#include<iostream> #include<cstdio> #include<cmath> #include<queue> #include<cstring> #include<queue> #include<algorithm> #include<map> #define ll long long using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f; } int n,ans=0; const int s[16]={1,0,0,0,1,0,1,0,2,1,1,2,0,1,0,0}; int main() { int x=read();if(!x)return 0*puts("1"); while(x) { int a=x%16; ans+=s[a]; x/=16; } cout<<ans; return 0; }
C.INTERCALC
当场就念了一首诗
DO YOU EXPECT ME TO FIND THIS OUT?
WHAT BASE AND/XOR LANGUAGE INCLUDES string?
DON'T BYTE OF MORE THAN YOU CAN CHEW
YOU CAN ONLY DISTORT THE LARGEST OF MATHEMATICS SO FAR
SAYING "ABRACADABRA" WITHOUT A MAGIC AND WON'T DO YOU ANY GOOD
THE LAST STACK RUPTURES. ALL DIE. OH, THE EMBARRASSMENT!
I HAVE NO ARRAY AND I MUST SCREAM
ELEMENTS MAY NOT BE STORED IN WEST HYPERSPACE
我们发现它是藏头的，其实表达的是
FIND/FIGURE OR/XOR OF/OFF LARGEST/LAWS AND/WAND LAST/NEXT ARRAY/FILE ELEMENTS/VARIABLES
find xor of largest and last array elements
所以找出最大的和最后一个异或一下输出
#include<cstdio> #include<iostream> using namespace std; int n,x,mx=0; int main() { cin>>n; for(int i=1;i<=n;i++) cin>>x,mx=max(mx,x); cout<<(x^mx); return 0; }
D.Touchy-feely Plaindromes
根据样例和题目判断，最后的结论是根据盲文判断是否对称...
#include<cstdio> #include<iostream> #include<cstring> using namespace std; int s[]={8,-1,-1,-1,6,9,4,-1,0,5}; char st[12313]; int main() { scanf("%s",st+1);int n=strlen(st+1); for(int i=1,j=n;i<j;i++,j--) if(st[i]!=st[j]&&s[st[i]-'0']!=st[j]-'0') return 0*puts("No"); if(n&1){if(st[n/2+1]!='3'&&st[n/2+1]!='7')return 0*puts("No");} puts("YES"); return 0; }
E.Twisted Circuit
第二列从上到下表示异或、或、与和异或，然后直接算
#include<iostream> #include<cstdio> #include<cmath> #include<queue> #include<cstring> #include<queue> #include<algorithm> #include<map> #define ll long long using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f; } int s[15]; int main() { int a=read(),b=read(),c=read(),d=read(); cout<<(((a^b)&(c|d))^((b&c)|(a^d))); return 0; }
F.Crunching Numbers Just for You
题意：有n个顾客有n个数，你要把给定的n<=10个数排序好。
但是排序完出现rejected的情况，这时候看最后一句话
Remember, this is a very important feature, and you have to make sure the customers appreciate it!
所以顾客不想那么快结束，你要卡卡时间就行了。
#include<iostream> #include<cstdio> #include<cmath> #include<queue> #include<cstring> #include<queue> #include<algorithm> #include<ctime> #include<map> #define ll long long using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f; } int a[132123]; int main() { int n=read();time_t start = clock(); for(int i=1;i<=n;i++)a[i]=read(); sort(a+1,a+n+1); while((double)(clock()-start)/CLOCKS_PER_SEC<1.9); for(int i=1;i<=n;i++) cout<<a[i]<<" "; return 0; }
F.给定一个表达式求值，输出一份BrainFuck语言写的可以算出答案的代码。
先算出答案，然后按照BrainFuck语言的一套理论，+表示计数器+1，-表示计数器-1，.（点）表示输出计数器的asc码对应的字符。所以+48之后随便输出呗。
#include<cstdio> #include<iostream> #include<cstring> using namespace std; char st[1231000]; int op[1231231],num[12313],cnt=0; int calc(int l,int r) { int x=0; for(int i=l;i<=r;i++) x=x*10+st[i]-'0'; return x; } void work(int&th,int x) { int to=x+48; while(th>to)th--,printf("-"); while(th<to)th++,printf("+"); printf(". "); } int main() { scanf("%s",st+1);int pre=0; for(int i=1;st[i];i++) { if(st[i]=='+'||st[i]=='-') { op[++cnt]=(st[i]=='+')?1:2; num[cnt]=calc(pre+1,i-1); pre=i; } } num[++cnt]=calc(pre+1,strlen(st+1)); int x=num[1]; for(int i=1;i<cnt;i++) { if(op[i]==1)x=x+num[i+1]; else x=x-num[i+1]; } int th=0;cnt=0; if(!x) {work(th,0);return 0;} while(x) { num[++cnt]=x%10;x/=10; } for(int i=cnt;i;i--) work(th,num[i]); return 0; }