while

签到题,没什么可说的
cpp
#include<bits/stdc++.h>
using namespace std;
string s;
string t="while";
signed main()
{
cin>>s;
int cnt=0;
for(int i=0;i<5;i++)
{
if(s[i]!=t[i])
cnt++;
}
cout<<cnt<<"\n";
return 0;
}
token

思路:前缀和直接处理一遍即可
cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
int n;
int a[100005];
int pre[100005];
signed main()
{
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i],pre[i]=pre[i-1]+a[i];
int ans=0;
for(int i=1;i<=n;i++)
{
ans=max(ans,pre[i]-pre[max(i-10,0LL)]);
}
cout<<ans;
return 0;
}
小苯的逆序对和

思路:我们可以考虑用栈去存储,自栈底到栈顶单调递减,我们每次只将当前值和之前的栈顶元素去比较不断更新最大值即可
cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
int t;
int n;
int a[2000005];
void solve()
{
stack<int> s;
cin>>n;
int ans=0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
if(s.empty())
{
s.push(a[i]);
continue;
}
while(!s.empty()&&a[i]>s.top())
{
s.pop();
}
if(!s.empty())
ans=max(ans,a[i]+s.top());
s.push(a[i]);
}
cout<<ans<<"\n";
}
signed main()
{
cin>>t;
while(t--)
{
solve();
}
return 0;
}
数组4.0

思路:先排序一下,然后双指针去扫,我们去找一个a[r]<=a[l]+1的一个区间,如果区间内出现了两个值,那么就是0,否则就是r-l的贡献,然后再不断去判断即可
cpp
#include <bits/stdc++.h>
using namespace std;
#define int long long
int T;
int n;
int a[200005];
void solve() {
cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + 1 + n);
int cnt = 0, l = 0, r = -1;
int t = 0;
for(int i = 1; i <= n; i++)
{
int x = a[i];
if(x > r + 1)
{
cnt += (l == r) ? t : 1;
l = x;
t = 0;
}
r = x;
t++;
}
cnt += (l == r) ? t : 1;
cout << cnt - 2 << "\n";
}
signed main() {
cin >> T;
while(T--)
solve();
return 0;
}
小苯的因子查询

先来看数学推导吧,看完推导估计大家就会了
字很丑大家请见谅
然后直接写代码即可
cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
int t;
int mod=998244353;
int dp[1000005];
int fast(int a,int b)
{
int base=1;
while(b>0)
{
if(b%2==1)
base=base*a%mod;
a=a*a%mod;
b/=2;
}
return base;
}
void solve()
{
int n;
cin>>n;
cout<<fast(dp[n]+1,mod-2)%mod<<" ";
}
signed main()
{
int cnt=0;
for(int i=1;i<=1000000;i++)
{
int num=i;
while(num%2==0)
num/=2,cnt++;
dp[i]=cnt;
}
cin>>t;
while(t--)
solve();
return 0;
}