First of all introduce a user-friendly IDE invincible - vs code, the Black well-written tutorials, thanks thanks, attach a link portal .
Problem: A and the least common multiple of the greatest common divisor (NEFU OJ 1077)
Template title.
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
int a,b;
while(cin >> a >> b)
{
int ans;
ans=__gcd(a,b); //内置gcd真香
cout << ans << " " << a/ans*b << endl;
}
return 0;
}Problem: B See GCD (NEFU OJ 992)
The core idea: violence, b is the gcd, and b = c, c must be greater than b!.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
ios::sync_with_stdio(false);
int a,b;
while(cin >> a >> b)
{
for(int i=b+1;;i++)
{
if(__gcd(i,a)==b)
{
cout << i << endl;
break;
}
}
}
return 0;
}Problem: the greatest common divisor of the number of the plurality of C (NEFU OJ 764)
The core idea: two two numbers gcd, and the results over a set number of the next gcd.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
ios::sync_with_stdio(false);
LL a[15],n;
while(cin >> n)
{
LL ans;
for(int i=1;i<=n;i++)
cin >> a[i];
ans=__gcd(a[1],a[2]);
for(int i=3;i<=n;i++)
ans=__gcd(ans,a[i]);
cout << ans << endl;
}
return 0;
}Problem: least common multiple of the number of the plurality of D (NEFU OJ 765)
The core idea: with ProbleC, find lcm (a, b) = a / gcd (a, b) * b, burst values can be avoided.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
ios::sync_with_stdio(false);
LL a[15],n;
while(cin >> n)
{
LL ans;
for(int i=1;i<=n;i++)
cin >> a[i];
ans=a[1]/__gcd(a[1],a[2])*a[2];
for(int i=3;i<=n;i++)
ans=ans/__gcd(ans,a[i])*a[i];
cout << ans << endl;
}
return 0;
}Problem:E LCM&GCD（NEFU OJ 1411）
The core ideas: violence, (I wrote a classmate ran 7ms, I wrote 500ms, follow I will try to think of QAQ !!), I had also thought that violence will time out, he has been to gcd, lcm want to pay union, then wa, and I will continue to follow-up verification.
My idea of violence: the use of property from a to b loop, and add the judgment condition.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
ios::sync_with_stdio(0);
int t;
while(cin >> t)
{
LL x,y;
while(t--)
{
cin >> x >> y;
LL ans=0,tmp,tmp2;
tmp=x*y; //lcm*gcd的性质
for(LL i=x; i<=y; i++)
{
if(tmp%i==0) //判定是否能整除
{
tmp2=__gcd(i,tmp/i);
if(tmp2==x&&tmp/i<=y)//二次判断，是否符合gcd==x，lcm不用进行判断，因为tmp是x*y的乘积，可以自己去带入lcm的公式验证一下
ans++;
}
}
cout << ans << endl;
}
}
return 0;
}Problem: F cute gcd (NEFU OJ 1221) (but I really do not love him ah QAQ !!)
The core ideas: mathematical derivation, as shown visible.
#include<bits/stdc++.h>
using namespace std;
int t, a, b;
int main()
{
ios::sync_with_stdio(0);
while(scanf("%d", &t) != EOF)
{
while(t--)
{
scanf("%d%d", &a, &b);
printf("%d\n", a*a-2*__gcd(a, b)*b);
}
}
return 0;
}Problem: G Takagi students factor (NEFU OJ 1669)
The core ideas: factoring, specific issues in the code comments.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
LL a,b;
while(~scanf("%lld%lld",&a,&b))
{
LL sum=0;
int ans=__gcd(a,b); //计算他是因为gcd是取a，b的公共因子的
for(int i=1;i*i<=ans;i++) //计算一半就行
if(ans%i==0)
sum++;
if(sqrt(ans)*sqrt(ans)==ans) //中间的因子是不是sqrt()为整数
printf("%lld\n",2*sum-1); //若是整数，×2就会多计算一次,例子在代码的最后面
else
printf("%lld\n",2*sum);
}
return 0;
}
//比如16 32的公共因子={1,2,4,8,16},sqrt(gcd(16,32))==4,他只能算一次，不能重复计算。Problem: H Fast modulo exponentiation (NEFU OJ 601)
The core idea: template title.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL fastPow(LL a,LL n,LL c)
{
LL base=a;
LL res=1;
const LL mod=c;
while(n)
{
if(n&1)
res=(res*base)%mod;
base=(base*base)%mod;
n>>=1;
}
return res;
}
int main()
{
ios::sync_with_stdio(false);
LL a,b,c;
while(cin>> a >> b >> c)
{
LL ans;
ans=fastPow(a,b,c);
cout << ans << endl;
}
return 0;
}Problem: I Kurt math (NEFU OJ 1666)
The core idea: You can write about their first violence, and played several successive answers that he will be able to find the law.
Let us talk about how I think of: casual working 1 <= n <= 1e18, if you write recursive, apparently arrays can not be open to 1e18, so you're looking for the law.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL M=1e9+7;
LL fastPow(LL a,LL n) //快速幂取模的模板
{
LL base=a;
LL res=1;
const LL mod=M;
while(n)
{
if(n&1)
res=(res*base)%mod;
base=(base*base)%mod;
n>>=1;
}
return res;
}
int main()
{
ios::sync_with_stdio(false);
LL n;
while(cin >> n)
{
LL ans;
ans=fastPow(3,n);
cout << (2*ans)%M << endl;//别忘了再一次取模
}
return 0;
}Problem: The number of XOR J solutions of the equations (NEFU OJ 1834)
This question was originally ljw card giant guy out of our AK, Xing'an family let him down, QAQ.
The core idea: lunch with the students think whim, I'll follow-up to prove it, and update.
Konjac write a proof, I do not know right or wrong, please correct me if wrong.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int fastPow(int a,int n)
{
int base=a;
int res=1;
while(n)
{
if(n&1)
res*=base;
base*=base;
n>>=1;
}
return res;
}
int main()
{
ios::sync_with_stdio(0);
LL n;
while(cin >> n)
{
int sum=0,ans;
while(n)
{
if(n&1)
sum++;
n>>=1;
}
ans=fastPow(2,sum);
cout << ans << endl;
}
return 0;
}
