ACM 第十三天 - 爱码网

题目地址：https://odzkskevi.qnssl.com/415c275cb0a15fcb4ede21b8cb5297de?v=1533963116

A题代码：

 1 #include <bits/stdc++.h>
 2  
 3 using namespace std;
 4  
 5 int main()
 6 {
 7     double n,m;
 8     freopen("input.txt","r",stdin);
 9     freopen("output.txt","w",stdout);
10     while(cin>>n>>m){
11         cout<<max(ceil(n*2.0/m),2.0)<<endl;
12     }
13 }

D题代码：

方法一数学方法

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int pow2(int a,int b)
 4 {
 5     int ans =1;
 6     for(int i=1;i<=b;i++)
 7         ans=(ans*2)%(1000000007);
 8     return ans;
 9 }
10 int main()
11 {
12     freopen("input.txt","r",stdin);
13     freopen("output.txt","w",stdout);
14     string s;
15     cin>>s;
16     int ans=0;
17     for(int i=0;i<s.size();i++)
18     {
19         if((s[i]=='N'||s[i]=='S') &&(s[i+1]=='W'||s[i+1]=='E'))
20             ans=(ans+1)%1000000007;
21 
22     }
23     long long a1=(long long)pow2(2,ans)%(1000000007);
24     cout<<a1<<endl;
25 
26     return 0;
27 }

方法二递推思想

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define mod 1000000007
 4 int main()
 5 {
 6     freopen("input.txt","r",stdin);
 7     freopen("output.txt","w",stdout);
 8     char s[10010];
 9     cin>>s;
10     int a[10010];
11     int s1=strlen(s);
12     a[1]=a[0]=1;
13     for(int i=1;i<=s1;i++)
14     {
15         if((s[i-1]=='N'||s[i-1]=='S') &&(s[i]=='W'||s[i]=='E'))
16                 a[i+1]=(a[i]+a[i-1])%mod;
17         else
18                 a[i+1]=a[i];
19     }
20     cout<<a[s1]<<endl;
21     return 0;
22 }

G题：一个数是三个素数的乘积

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     freopen("input.txt","r",stdin);
 7     freopen("output.txt","w",stdout);
 8     int n;
 9     bool flag;
10     while(cin>>n)
11     {
12         int cnt=0;
13         for(int i=2; i*i<=n; i++)
14         {
15             flag=1;
16             while(n%i==0)
17             {
18                 if(flag)
19                 {
20                     cnt++;
21                     flag=0;
22                 }
23                 n/=i;
24             }
25         }
26         if(n!=1) cnt++;
27         if(cnt==3) cout<<"YES"<<endl;
28         else cout<<"NO"<<endl;
29     }
30     return 0;
31 }