牛客周赛 Round 91

while

签到题，没什么可说的

#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

#include<bits/stdc++.h>
using namespace std;
#define int long longint 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

#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

#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 

#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;
}