1:Pell数列

#include <iostream> using namespace std; int main() { int n; long k; long long P[1000000+1]={0}; P[1]=1; P[2]=2; for (long i=3;i<=1000000; i++) P[i] = (2 * P[i-1] + P[i-2])%32767; cin >> n; while (n--) { cin >> k; cout << P[k] << endl; } return 0; }

2:求最大公约数问题

#include <iostream> using namespace std; long long gy(long long a, long long b) { if (a%b==0) return b; return gy(b, a%b); } int main() { long long a,b; cin >> a >> b; cout << gy(a, b) << endl; return 0; }

3:编程填空：第i位替换

#include <iostream> using namespace std; int bitManipulation1(int n, int m, int i) { // 在此处补充你的代码 return (((m>>i)&1) == 0) ? (n & (~(1<<i))) : (n | (1<<i)); } int main() { int n, m, i, t; cin >> t; while (t--) { cin >> n >> m >> i; cout << bitManipulation1(n, m, i) << endl; } return 0; }

4:编程填空：第i位取反

#include <iostream> using namespace std; int bitManipulation2(int n, int i) { // 在此处补充你的代码 return (((n>>i)&1)==0) ? (n | 1<<i) : (n & (~(1<<i))); } int main() { int t, n, i; cin >> t; while (t--) { cin >> n >> i; cout << bitManipulation2(n, i) << endl; } return 0; }

5:编程填空：左边i位取反

#include <iostream> #include <bitset> using namespace std; int bitManipulation3(int n, int i) { // 在此处补充你的代码 cout << bitset<sizeof(int)*8>(n) << endl; cout << bitset<sizeof(int)*8>((unsigned int)n>>(sizeof(n)*8-i)) << endl; cout << bitset<sizeof(int)*8>(~(n>>(sizeof(n)*8-i))) << endl; cout << bitset<sizeof(int)*8>((~(n>>(sizeof(n)*8-i)))<<(sizeof(n)*8-i)) << endl; cout << endl; unsigned long m = (unsigned int)n; int o = (int)(((m<<i)&(0x0ffffffff))>>i); cout << n << " " << i << " " << m << endl; cout << bitset<sizeof(int)*8>(n<<i) << endl; cout << bitset<sizeof(int)*8>(m<<i) << endl; cout << bitset<sizeof(int)*8>(o) << endl; cout << bitset<sizeof(int)*8>( ((~(n>>(sizeof(n)*8-i)))<<(sizeof(n)*8-i)) + (int)((((unsigned long)n<<i)&(0x0ffffffff))>>i) )<< endl; return ((~(n>>(sizeof(n)*8-i)))<<(sizeof(n)*8-i)) + (int)((((unsigned long)n<<i)&(0x0ffffffff))>>i); } int main() { int t, n, i; cin >> t; while (t--) { cin >> n >> i; cout << bitManipulation3(n, i) << endl; } return 0; }

6:编程填空：计算整数k

#include <iostream> #include <iomanip> using namespace std; int main() { unsigned int t, n, i, j, m, mi, mj; cin >> t; while (t--) { cin >> n >> i >> j; mi = (n>>i)&1; mj = (((n>>j)&1)==0)?1:0; m = (mi<<i) + (mj<<j); for (unsigned int k=i+1; k<j; k++) m += 1<<k; cout << setbase(16) << m << endl; } return 0; }

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