A. Numbers Joke

time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output Input

The input contains a single integer a (1?≤?a?≤?30).

Output

Output a single integer.

Example Input 3 Output 27

題目鏈接：http://codeforces.com/contest/784/problem/A

分析：百度史蒂芬數，直接打表求解！

下面給出AC代碼：

1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 int n; 6 while(cin>>n) 7 { 8 if(n==1) 9 cout<<4<<endl; 10 else if(n==2) 11 cout<<22<<endl; 12 else if(n==3) 13 cout<<27<<endl; 14 else if(n==4) 15 cout<<58<<endl; 16 else if(n==5) 17 cout<<85<<endl; 18 else if(n==6) 19 cout<<94<<endl; 20 else if(n==7) 21 cout<<121<<endl; 22 else if(n==8) 23 cout<<166<<endl; 24 else if(n==9) 25 cout<<202<<endl; 26 else if(n==10) 27 cout<<265<<endl; 28 else if(n==11) 29 cout<<274<<endl; 30 else if(n==12) 31 cout<<319<<endl; 32 else if(n==13) 33 cout<<346<<endl; 34 else if(n==14) 35 cout<<355<<endl; 36 else if(n==15) 37 cout<<378<<endl; 38 else if(n==16) 39 cout<<382<<endl; 40 else if(n==17) 41 cout<<391<<endl; 42 else if(n==18) 43 cout<<438<<endl; 44 else if(n==19) 45 cout<<454<<endl; 46 else if(n==20) 47 cout<<483<<endl; 48 else if(n==21) 49 cout<<517<<endl; 50 else if(n==22) 51 cout<<526<<endl; 52 else if(n==23) 53 cout<<535<<endl; 54 else if(n==24) 55 cout<<562<<endl; 56 else if(n==25) 57 cout<<576<<endl; 58 else if(n==26) 59 cout<<588<<endl; 60 else if(n==27) 61 cout<<627<<endl; 62 else if(n==28) 63 cout<<634<<endl; 64 else if(n==29) 65 cout<<636<<endl; 66 else if(n==30) 67 cout<<645<<endl; 68 } 69 return 0; 70 }

B. Kids' Riddle

time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output

Programmers' kids solve this riddle in 5-10 minutes. How fast can you do it?

Input

The input contains a single integer n (0?≤?n?≤?2000000000).

Output

Output a single integer.

Examples Input 11 Output 2 Input 14 Output 0 Input 61441 Output 2 Input 571576 Output 10 Input 2128506 Output 3 題目鏈接：http://codeforces.com/contest/784/problem/B 分析：化成16進制，然后數圈圈(B有兩個洞。。。。。) 下面給出AC代碼： 1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 string s; 5 int a,b,c,d; 6 int main() 7 { 8 int ans; 9 cin>>ans; 10 if(ans==0) 11 { 12 a++; 13 } 14 while(ans) 15 { 16 if(ans%16==10) 17 { 18 a++; 19 } 20 else if(ans%16==11) 21 { 22 a+=2; 23 } 24 else if(ans%16==13) 25 { 26 a++; 27 } 28 else if(ans%16==6) 29 { 30 a++; 31 } 32 else if(ans%16==8) 33 { 34 a+=2; 35 } 36 else if(ans%16==9) 37 { 38 a++; 39 } 40 else if(ans%16==0) 41 { 42 a++; 43 } 44 else if(ans%16==4) 45 { 46 a++; 47 } 48 ans/=16; 49 } 50 cout<<a<<endl; 51 return 0; 52 }

C. INTERCALC

time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output

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

Input

The first line of input data contains a single integer n (1?≤?n?≤?10).

The second line of input data contains n space-separated integers a_i (1?≤?a_i?≤?11).

Output

Output a single integer.

Example Input 4
2 5 3 1 Output 4 ?????????????? 題目鏈接：http://codeforces.com/contest/784/problem/C 分析：把每句其中幾個字取出來，組合在一起，就是最大值和最后一個數異或 下面給出AC代碼： 1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 int n,x,mx=0; 5 int main() 6 { 7 cin>>n; 8 for(int i=1;i<=n;i++) 9 cin>>x,mx=max(mx,x); 10 cout<<(x^mx); 11 return 0; 12 }

D. Touchy-Feely Palindromes

time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output Input

The only line of the input contains a string of digits. The length of the string is between 1 and 10, inclusive.

Output

Output "Yes" or "No".

Examples Input 373 Output Yes Input 121 Output No Input 436 Output Yes

題目鏈接：http://codeforces.com/contest/784/problem/D

分析：看是不是回文（436中4,6盲文對稱,所以s[4]=6）

下面給出AC代碼： 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 using namespace std; 5 6 int s[]={8,-1,-1,-1,6,9,4,-1,0,5}; 7 char st[12313]; 8 9 int main() 10 { 11 scanf("%s",st+1);int n=strlen(st+1); 12 for(int i=1,j=n;i<j;i++,j--) 13 if(st[i]!=st[j]&&s[st[i]-'0']!=st[j]-'0') return 0*puts("No"); 14 if(n&1){if(st[n/2+1]!='3'&&st[n/2+1]!='7')return 0*puts("No");} 15 puts("YES"); 16 return 0; 17 }

E. Twisted Circuit

time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output Input

The input consists of four lines, each line containing a single digit 0 or 1.

Output

Output a single digit, 0 or 1.

Example Input 0
1
1
0 Output 0

題目鏈接：http://codeforces.com/contest/784/problem/E

分析：一個數字電路，不過或門和異或門是反的，所以。。注意下

下面給出AC代碼：

1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 string s; 5 int a,b,c,d; 6 int main() 7 { 8 cin>>a>>b>>c>>d; 9 printf("%d\n",((a^b)&(c|d))^((b&c)|(a^d))); 10 return 0; 11 }

F. Crunching Numbers Just for You

time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output

You are developing a new feature for the website which sells airline tickets: being able to sort tickets by price! You have already extracted the tickets' prices, so there's just the last step to be done...

You are given an array of integers. Sort it in non-descending order.

Input

The input consists of a single line of space-separated integers. The first number is n (1?≤?n?≤?10) — the size of the array. The following n numbers are the elements of the array (1?≤?a_i?≤?100).

Output

Output space-separated elements of the sorted array.

Example Input 3
3 1 2 Output 1 2 3 Note

Remember, this is a very important feature, and you have to make sure the customers appreciate it!

題目鏈接：http://codeforces.com/contest/784/problem/F

分析：

排序。不過必須運行時間超過1s，不會怎么控制時間？有個好辦法，隨便找個代碼，反正要運行1s以上的，加進去，然后輸出結果就好了，這里我用了網絡賽的代碼

下面給出AC代碼：

1 #include<bits/stdc++.h> 2 using namespace std; 3 4 #define MAXN 100 5 #define MAXM 10001 6 #define MAXP 266666 7 #define MAX 3200001 8 #define clr(ar) memset(ar, 0, sizeof(ar)) 9 #define read() freopen("lol.txt", "r", stdin) 10 #define dbg(x) cout << #x << " = " << x << endl 11 #define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31)))) 12 #define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31)))) 13 #define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2)) 14 15 16 namespace pcf 17 { 18 long long dp[MAXN][MAXM]; 19 unsigned int ar[(MAX >> 6) + 5] = {0}; 20 int len = 0, primes[MAXP], counter[MAX]; 21 22 void Sieve() 23 { 24 setbit(ar, 0), setbit(ar, 1); 25 for (int i = 3; (i * i) < MAX; i++, i++) 26 { 27 if (!chkbit(ar, i)) 28 { 29 int k = i << 1; 30 for (int j = (i * i); j < MAX; j += k) setbit(ar, j); 31 } 32 } 33 34 for (int i = 1; i < MAX; i++) 35 { 36 counter[i] = counter[i - 1]; 37 if (isprime(i)) primes[len++] = i, counter[i]++; 38 } 39 } 40 41 void init() 42 { 43 Sieve(); 44 for (int n = 0; n < MAXN; n++) 45 { 46 for (int m = 0; m < MAXM; m++) 47 { 48 if (!n) dp[n][m] = m; 49 else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]]; 50 } 51 } 52 } 53 54 long long phi(long long m, int n) 55 { 56 if (n == 0) return m; 57 if (primes[n - 1] >= m) return 1; 58 if (m < MAXM && n < MAXN) return dp[n][m]; 59 return phi(m, n - 1) - phi(m / primes[n - 1], n - 1); 60 } 61 62 long long Lehmer(long long m) 63 { 64 if (m < MAX) return counter[m]; 65 66 long long w, res = 0; 67 int i, a, s, c, x, y; 68 s = sqrt(0.9 + m), y = c = cbrt(0.9 + m); 69 a = counter[y], res = phi(m, a) + a - 1; 70 for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1; 71 return res; 72 } 73 } 74 75 long long solve(long long n) 76 { 77 int i, j, k, l; 78 long long x, y, res = 0; 79 80 /*for (i = 0; i < pcf::len; i++){ 81 printf("%I64d\n",pcf::Lehmer(n)); 82 x = pcf::primes[i], y = n / x; 83 if ((x * x) > n) break; 84 res += (pcf::Lehmer(y) - pcf::Lehmer(x)); 85 } 86 87 for (i = 0; i < pcf::len; i++){ 88 x = pcf::primes[i]; 89 if ((x * x * x) > n) break; 90 res++; 91 }*/ 92 res=pcf::Lehmer(n); 93 return res; 94 } 95 int xx[100]; 96 int main() 97 { 98 pcf::init(); 99 long long n, res; 100 while(cin>>n) 101 { 102 int x=solve(100000000000); 103 for(int i=1; i<=n; i++) 104 { 105 cin>>xx[i]; 106 } 107 sort(xx+1,xx+n+1); 108 for(int i=1; i<=n; i++) 109 { 110 cout<<xx[i]<<" "; 111 } 112 } 113 return 0; 114 }

G. BF Calculator

time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output

In this problem you will write a simple generator of Brainfuck (https://en.wikipedia.org/wiki/Brainfuck) calculators.

You are given an arithmetic expression consisting of integers from 0 to 255 and addition/subtraction signs between them. Output a Brainfuck program which, when executed, will print the result of evaluating this expression.

We use a fairly standard Brainfuck interpreter for checking the programs:

• 30000 memory cells.
• memory cells store integers from 0 to 255 with unsigned 8-bit wraparound.
• console input (, command) is not supported, but it's not needed for this problem.

Input

The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive. The calculations result is guaranteed to be an integer between 0 and 255, inclusive (results of intermediary calculations might be outside of these boundaries).

Output

Output a Brainfuck program which, when executed, will print the result of evaluating this expression. The program must be at most 5000000 characters long (including the non-command characters), and its execution must be complete in at most 50000000 steps.

Examples Input 2+3 Output ++>
+++>
<[<+>-]<
++++++++++++++++++++++++++++++++++++++++++++++++. Input 9-7 Output +++++++++>
+++++++>
<[<->-]<
++++++++++++++++++++++++++++++++++++++++++++++++. Note

You can download the source code of the Brainfuck interpreter by the link http://assets.codeforces.com/rounds/784/bf.cpp. We use this code to interpret outputs.

題目鏈接：http://codeforces.com/contest/784/problem/G 分析：

給定一個表達式求值，輸出一份BrainFuck語言寫的可以算出答案的代碼。

先算出答案，然后按照BrainFuck語言的一套理論，+表示計數器+1，-表示計數器-1，.（點）表示輸出計數器的asc碼對應的字符。所以+48之后隨便輸出唄！

下面給出AC代碼：

1 #include <bits/stdc++.h> 2 using namespace std; 3 4 char st[1231000]; 5 int op[1231231],num[12313],cnt=0; 6 7 int calc(int l,int r) 8 { 9 int x=0; 10 for(int i=l;i<=r;i++) 11 x=x*10+st[i]-'0'; 12 return x; 13 } 14 15 void work(int&th,int x) 16 { 17 int to=x+48; 18 while(th>to)th--,printf("-"); 19 while(th<to)th++,printf("+"); 20 printf(".\n"); 21 } 22 23 int main() 24 { 25 scanf("%s",st+1);int pre=0; 26 for(int i=1;st[i];i++) 27 { 28 if(st[i]=='+'||st[i]=='-') 29 { 30 op[++cnt]=(st[i]=='+')?1:2; 31 num[cnt]=calc(pre+1,i-1); 32 pre=i; 33 } 34 } 35 num[++cnt]=calc(pre+1,strlen(st+1)); 36 int x=num[1]; 37 for(int i=1;i<cnt;i++) 38 { 39 if(op[i]==1)x=x+num[i+1]; 40 else x=x-num[i+1]; 41 } 42 int th=0;cnt=0; 43 if(!x) {work(th,0);return 0;} 44 while(x) 45 { 46 num[++cnt]=x%10;x/=10; 47 } 48 for(int i=cnt;i;i--) 49 work(th,num[i]); 50 return 0; 51 }

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總結

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