程序设计与算法 第六周测验

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<
4000
/span>(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;
}