• cf 764A、762A、764B


     颓废题

    764A

     1 #include<bits/stdc++.h> 
     2 #define LL long long 
     3 #define N 100005
     4 #define lowbit(x) x&(-x)
     5 using namespace std;
     6 inline int ra()
     7 {
     8     int x=0,f=1; char ch=getchar();
     9     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    11     return x*f;
    12 }
    13 int main()
    14 {
    15     int n=ra(),m=ra(),z=ra();
    16     cout<<z/(n*m/__gcd(n,m));
    17     return 0;
    18 }

    764B

     1 #include<bits/stdc++.h> 
     2 #define LL long long 
     3 #define N 100005
     4 #define lowbit(x) x&(-x)
     5 using namespace std;
     6 inline int ra()
     7 {
     8     int x=0,f=1; char ch=getchar();
     9     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    11     return x*f;
    12 }
    13 int a[N<<1];    
    14 int main()
    15 {
    16     int n=ra();
    17     for (int i=1; i<=n; i++) a[i]=ra();
    18     for (int i=1; i<=n/2; i++)
    19     {
    20         if (i%2) swap(a[i],a[n-i+1]);
    21     }
    22     for (int i=1; i<=n; i++)
    23         printf("%d ",a[i]);
    24     return 0;
    25 }

    762A    (这道题简直做弱智了。。。)

     1 #include<bits/stdc++.h> 
     2 #define LL long long 
     3 #define N 100005
     4 #define lowbit(x) x&(-x)
     5 using namespace std;
     6 inline int ra()
     7 {
     8     int x=0,f=1; char ch=getchar();
     9     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    11     return x*f;
    12 }
    13 int a[N<<1];
    14 int cnt;
    15 int main()
    16 {
    17     LL n; cin>>n; int k=ra();
    18     for (int i=1; i<=(int)sqrt(n); i++)
    19     {
    20         if (n%(LL)i==0) a[++cnt]=i;
    21     }
    22     if (k<=cnt) cout<<a[k];
    23     else {
    24         if (n!=(LL)sqrt(n)*(LL)sqrt(n))
    25         {
    26             if (k>cnt*2) cout<<"-1";
    27             else cout<<(n/(LL)a[cnt*2+1-k]);
    28         }
    29         else
    30         {
    31             if (k>cnt*2-1) cout<<"-1";
    32             else cout<<(n/(LL)a[cnt*2-k]);
    33         }
    34     }
    35     return 0;
    36 }
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  • 原文地址:https://www.cnblogs.com/ccd2333/p/6364515.html
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