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一、1.求阶乘 - 蓝桥云课

算法代码：

#include <bits/stdc++.h>
using namespace std;
#define  ll long long
ll check(ll n)
{ll cnt=0;while(n){cnt+=(n/=5);}return cnt;
}int main()
{ll k;cin>>k;ll L=0,R=1e19;while(L<R){ll mid=(L+R)>>1;if(check(mid)>=k){R=mid;}else{L=mid+1;}}  if(check(R)==k){cout<<R;}else{cout<<-1;}return 0;
}

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二、1.青蛙过河 - 蓝桥云课

 

算法代码： 

#include <bits/stdc++.h>
using namespace std;
int n,x;
int h[100005];
int sum[100005];bool check(int mid)
{for(int i=1;i<n-mid+1;i++){if(sum[i+mid-1]-sum[i-1]<2*x){return false;}}return true;
}int main()
{cin>>n>>x;sum[0]=0;for(int i=1;i<n;i++){cin>>h[i];sum[i]=sum[i-1]+h[i];}int L=1,R=n;while(L<R){int mid=(L+R)/2;if(check(mid)){R=mid;}else{L=mid+1;}}cout<<L;return 0;
}

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三、1.管道 - 蓝桥云课

算法代码：

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
const int LEN=1e9;
int n,len;
int L[N],S[N];bool check(int t)
{int cnt=0;int last_L=2,last_R=1;for(int i=0;i<n;i++){if(t>=S[i]){cnt++;int left=L[i]-(t-S[i]);int right=L[i]+(t-S[i]);if(left<last_L){last_L=left,last_R=max(last_R,right);}else if(left<=last_R+1){last_R=max(last_R,right);}}}if(cnt==0){return false;}if(last_L<=1&&last_R>=len){return true;}else{return false;}
}int main()
{scanf("%d%d",&n,&len);for(int i=0;i<n;i++){scanf("%d%d",&L[i],&S[i]);}int LL=0,R=2e9,ans=-1;while(LL<=R){int mid=((R-LL)>>1)+LL;if(check(mid)){ans=mid,R=mid-1;}else{LL=mid+1;}}printf("%d\n",ans);return 0;
}

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 四、1.技能升级 - 蓝桥云课

算法代码： 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;    // Note: long long is needed
const int N = 100100;
int a[N], b[N];    // Store a_i, b_i
int n, m;bool check(ll mid) {    // Check if the last skill upgrade can reach midll cnt = 0;for (int i = 0; i < n; ++i) {if (a[i] < mid)continue;    // Initial value of skill i is less than mid, skipcnt += (a[i] - mid) / b[i] + 1;    // Number of times skill i is usedif (cnt >= m)    // Total upgrades ≥ m, mid is too smallreturn true;}return false;    // Total upgrades < m, mid is too large
}int main() {cin >> n >> m;for (int i = 0; i < n; ++i)cin >> a[i] >> b[i];ll L = 1, R = 1000000;    // Binary search for the highest possible last attackwhile (L <= R) {ll mid = (L + R) / 2;if (check(mid)) L = mid + 1;    // Increase midelse R = mid - 1;    // Decrease mid}ll attack = 0;ll cnt = m;for (int i = 0; i < n; ++i) {if (a[i] < R) continue;ll t = (a[i] - L) / b[i] + 1;    // Number of upgrades for skill iif (a[i] - b[i] * (t - 1) == R)t -= 1;    // If skill's upgrade equals R exactly, other skills are betterattack += (a[i] * 2 - (t - 1) * b[i]) * t / 2;cnt -= t;}cout << attack + cnt * R << endl;return 0;
}