Ab two numbers are given to you ask Feibolaqi number and a column item to the second item of b is odd or even
Let's take a concept of parity constitute 察斐波拉契 series of cycles even Kiki Kiki Kiki even ...... even such a group of three
Then we were the first to determine ab a few number of cycles to keep on with a string and each character b is then converted to an integer corresponding to each digit is -48
With each number to be able to get 3 remainder Which one is the number of cycles in the
Then see below
(Simple push came out)
AC Code:
1 #include<bits/stdc++.h> 2 #define pi acos(-1) 3 typedef long long ll; 4 typedef unsigned long long ull; 5 using namespace std; 6 7 namespace io { 8 const int SIZE = 1e7 + 10; 9 char inbuff[SIZE]; 10 char *l, *r; 11 inline void init() { 12 l = inbuff; 13 r = inbuff + fread(inbuff, 1, SIZE, stdin); 14 } 15 inline char gc() { 16 if (l == r) init(); 17 return (l != r) ? *(l++) : EOF; 18 } 19 void read(int &x) { 20 x = 0; char ch = gc(); 21 while(!isdigit(ch)) ch = gc(); 22 while(isdigit(ch)) x = x * 10 + ch - '0', ch = gc(); 23 } 24 } using io::read; 25 26 bool cmp(const int &a, const int &b){ 27 return a > b; 28 } 29 30 int main(){ 31 ios::sync_with_stdio(false); 32 int t; 33 cin>>t; 34 string a, b; 35 while (t--){ 36 cin>>a>>b; 37 int sum1 = 0, sum2 = 0; 38 for (int i = 0; i < a.size(); i++) 39 sum1 += int(a[i]) - 48; 40 for (int i = 0; i < b.size(); i++) 41 sum2 += int(b[i]) - 48; 42 int tmp1 = sum1 % 3, tmp2 = sum2 % 3; 43 if ((tmp1 == 0 && tmp2 == 1) || (tmp1 == 1 && tmp2 == 1) || 44 (tmp1 == 2 && tmp2 == 0) || (tmp1 == 2 && tmp2 == 2)) 45 cout<<1<<endl; 46 else cout<<0<<endl; 47 } 48 return 0; 49 }