A
简单减法
代码实现
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| #include<bits/stdc++.h>
using namespace std;
#define ll long long
int main(){
int l,r;
l=r=0;
string s;
int n;
scanf("%d",&n);
cin>>s;
for(int i=0;i<n;i++)
{
if(s[i]=='L')l--;
else r++;
}
printf("%d\n",r-l+1);
}
|
B
思路
从0-n-1,1-n找最大的区间和即可
代码实现
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| #include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5+10;
int a[N];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{ bool flag = false;
bool all = true;
int n;
scanf("%d",&n);
ll temp = 0;
ll sum = 0;
ll maxsum = -1e18;
for(int i=0;i<n;i++)scanf("%d",&a[i]),sum+=a[i];
for(int i=1;i<n;i++)
{
if(a[i]<=0)flag=true;
temp+=a[i];
if(temp<0)temp=0,all=false;
maxsum = max(maxsum,temp);
}
temp=0;
for(int i=0;i<n-1;i++)
{
if(a[i]<=0)flag=true;
temp+=a[i];
if(temp<0)temp=0,all=false;
maxsum = max(maxsum,temp);
}
if(!flag)
{
printf("YES\n");
continue;
}
if(maxsum>=sum)printf("NO\n");
else printf("YES\n");
}
}
|
C
思路
直接暴力sqrt(n)枚举，判断即可
代码实现
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| #include<bits/stdc++.h>
using namespace std;
#define ll long long
ll gcd(ll a,ll b)
{
return b?gcd(b,a%b):a;
}
int main()
{
ll x;
scanf("%lld",&x);
ll a=1,b=x;
ll maxx = x;
ll sx = ceil(sqrt(x));
for(int i=2;i<sx;i++)
{
if(x%i==0&&(x%(1ll*i*i)!=0)){
ll aa = i;
ll bb = x/i;
ll temp = aa*bb/gcd(aa,bb);
if(temp==x&&max(aa,bb)<maxx)a=aa,b=bb;
}
}
printf("%lld %lld\n",a,b);
}
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D
思路
把数字转换为二进制之后一列一列的看，如果这一列的0与1小于等于1个的话那么就加上(1<<bits)，如果为0则不管。
递归的往下搜索即可
代码实现
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| #include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5+10;
bool vis[N];
vector<int>a;
ll solve(vector<int> &c,int bits)
{
if(bits<0)return 0;
vector<int>l,r;
for(auto it : c)
{
if(((1<<bits)&it)==0)l.push_back(it);
else r.push_back(it);
}
if(l.size()==0)return solve(r,bits-1);
if(r.size()==0)return solve(l,bits-1);
return min(solve(l,bits-1),solve(r,bits-1))+(1<<bits);
}
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
int x;
scanf("%d",&x);
a.push_back(x);
}
printf("%lld\n",solve(a,29));
}
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