1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <algorithm>
 8 #include <string>
 9 #include <queue>
10 #include <map>
11 #include <vector>
12
 
 #include <iomanip>
13 using namespace std;
14 const int maxn = 1e3+50;
15 const int maxm = 1e4+10;
16 const int inf = 0x3f3f3f3f;
17 const double epx = 1e-10;
18 typedef long long ll;
19 const ll INF = 1e18;
20 const double pi = acos(-1.0);
21 int main()
22 {
23     int t,c;
24     double a,b;
25     cin>>t;
 
26     while(t--)
27     {
28         cin>>a>>b>>c;
29         cout<<setiosflags(ios::fixed)<<setprecision(c)<<(double)pow(a,exp(1))/b<<endl;//控制輸出小數點後位數
30     }
31 }

I題 卡特蘭數 出棧序列有多少種

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string
 
.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <algorithm>
 8 #include <string>
 9 #include <queue>
10 #include <map>
11 #include <vector>
12 #include <iomanip>
13 #define mem(a,b) memset(a,b,sizeof(a))
14 using namespace std;
15 const int maxn = 5e5+50;
16 const int inf = 0x3f3f3f3f;
17 const double epx = 1e-10;
18 const double pi = acos(-1.0);
19 typedef long long ll;
20 const ll INF = 1e18;
21 const ll mod = 998244353;
22 ll n,f[maxn];
23 void ff()
24 {
25     f[0]=1;
26     for(int i=1;i<=500000;i++)
27         f[i]=(i*f[i-1])%mod;
28 }
29 ll poww(ll n,ll m)
30 {
31     ll ans = 1;
32     while(m > 0)
33     {
34         if(m & 1)ans = (ans * n) % mod;
35         m = m >> 1;
36         n = (n * n) % mod;
37     }
38     return ans;
39 }
40 ll cc(ll n,ll m)
41 {
42     ll ans=f[n];
43     ans=(ans*poww(f[m],mod-2))%mod;
44     ans=(ans*poww(f[n-m],mod-2))%mod;
45     return ans;
46 }
47 int main()
48 {
49     int t,kase=1;
50     cin>>t;
51     ff();
52     //cout<<cc(6,3)<<endl;
53     while(t--)
54     {
55         cin>>n;
56         int m=n-1;
57         ll ans1=(cc(2*n,n)-cc(2*n,n+1))%mod;
58         ll ans2=(cc(2*m,m)-cc(2*m,m+1))%mod;
59         //cout<<ans1<<" "<<ans2<<endl;
60         printf("Case #%d: %lld\n",kase++,(ans1-ans2+2*mod)%mod);
61     }
62 }

牛客網小白月賽1 B，I