Codeforces Round #195 (Div. 2)

记次CF吧 1题。。。B题。。因为循环的i没设成long long 却参与了运算 结果就悲剧了 一直交 一直挂 。。上题

A 水。。 第一次少了个空格还。。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<stdlib.h>
 #include<algorithm>
 using namespace std;
 #define LL long long
 LL x,y;
 int main()
 {
     int i,j,k,n,m;
     cin>>x>>y;
     if((x<&&y>)||(x>&&y<))
     {
         LL b = y-x;
         if(b<)
         cout<<""<<" "<<b<<" "<<-b<<" "<<""<<endl;
         else
         cout<<-b<<" "<<"0 "<<""<<" "<<b<<endl;
     }
     else
     {
         LL b = y+x;
         if(b>)
         cout<<""<<" "<<b<<" "<<b<<" "<<""<<endl;
         else
         cout<<b<<" "<<""<<" "<<"0 "<<b<<endl;
     }
     return ;
 }

B题 大体画画 就出来了 求下和加加

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<cmath>
 using namespace std;
 #define eps 1e-6
 int main()
 {
     long long m,r,i;
     cin>>m>>r;
     double s =,s1=;
     for(i =  ; i <= m ; i++)
     {
         s = ;
         if(i>)
         s+=(i-)*(i-)*r+(i-2.0)*sqrt(2.0)**r+sqrt(2.0)*r+*r;
         else
         s+=(i-)*i*r+(i-)*sqrt(2.0)*r;
         if(m-i>=)
         s+=(m-i-)*(m-i)*r+(m-i-)*sqrt(2.0)**r+*r+sqrt(2.0)*r+*r;
         else
         s+=(m-i)*(m-i+)*r+(m-i)*sqrt(2.0)*r+*r;
         s1+=s/m;
     }
     printf("%.10f\n",s1/m);
     return ;
 }

往后就没再看 一直在交B 7次WA啊啊 泪~~

补道C题

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #define N 100010
 #define LL long long
 using namespace std;
 struct node
 {
     LL a;
     int d[],k;
 }q[N];
 int f[];
 void digit(int e)
 {
     int g=;
     LL y = q[e].a;
     while(y)
     {
         int x = y%;
         g++;
         q[e].d[g] = x;
         y/=;
     }
     q[e].k = g;
 }
 bool cmp(node a,node b)
 {
     return a.a<b.a;
 }
 int main()
 {
     int i,j,k,n,o,g;
     cin>>n;
     for(i = ; i <= n ;i++)
     {
         scanf("%d",&q[i].a);
         digit(i);
     }
     sort(q+,q+n+,cmp);
     o = ;
     for(i = ; i >=  ; i--)
     {
         memset(f,,sizeof(f));
         for(j = n; j >=  ; j--)
         {
             if(q[j].k<i)
             break;
             if(q[j].d[i]!=)
             {
                 for(g = ; g < q[j].k ; g++)
                 if(q[j].d[g]==)
                     f[g] = ;
             }
         }
         for(g =  ; g < i ; g++)
             if(!f[g]) break;
         if(g==i)
         {
             o = g;
             break;
         }
     }
     int num = ;
     for(i = n ; i >=  ; i--)
     if(q[i].d[o]==)
     num++;
     cout<<num<<endl;
     int w=;
     for(i = n ; i >=  ; i--)
     if(q[i].d[o]==)
     {
         if(w)
         printf(" ");
         w++;
         cout<<q[i].a;
     }
     return ;
 }

补道D题

根据费马小定理神马的求逆元 然后高端的算组合数取模

若第一个是1 则最后为0  若全是0且为偶数-》1 否则-》0 组合起来就可以求解了

100。。。。

0100。。

0010。。

0001。。。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 using namespace std;
 #define mod 1000000007
 #define LL long long
 LL ff[];
 LL fmod(LL a,LL k)
 {
     LL b = ;
     while(k)
     {
         if(k&)
         b = a*b%mod;
         a = (a%mod)*(a%mod)%mod;
         k/=;
     }
     return b;
 }
 LL cn(int n,int m)
 {
     LL ans,a;
     ans = ff[n];
     a = fmod((ff[n-m]*ff[m])%mod,mod-);
     return (ans*a)%mod;
 }
 int main()
 {
     int i,j,k,n,m,g;
     while(cin>>n>>m>>g)
     {
         LL s1=,s2=;
         if(n==)
         {
             if(m==&&g==)
             cout<<"1\n";
             else if(m>&&g==)
             cout<<"0\n";
             else if(m==&&g==)
             cout<<"0\n";
             else
             cout<<"1\n";
             continue;
         }
         else if(m==)
         {
             if(n%==)
             k = ;
             else
             k = ;
             if(k==g)
             cout<<"1\n";
             else
             cout<<"0\n";
             continue;
         }
         ff[] = ;
         for(i = ;i <= n+m;i ++)
         {
            ff[i] = (ff[i-]*i)%mod;
         }
         s1 = cn(n+m,m);
         for(i =  ;i <= n ; i+=)
         {
             if(m+n-i->=m-)
             s2=(s2+cn(m+n-i-,m-))%mod;
         }
         if(m==&&n%!=)
         s2++;
         if(m==&&n%==)
         s2--;
         if(g==)
         cout<<s2<<endl;
         else
         {
             if(s1-s2<)
             s1 = s1-s2+mod;
             else
             s1 = s1-s2;
             cout<<s1<<endl;
         }
     }
     return ;
 }