牛客练习赛32 A/B/C

https://ac.nowcoder.com/acm/contest/272/A

v<=k时  答案就是k个1

否则贪心的从中间向两边添加1

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define pii pair<int,int>
 #define inf 0x3f3f3f3f
 #define pb push_back
 #define mp make_pair
 const LL mod=1e9+;
 int s[];
 int main(){
    int v,k,i,j;
    while(cin>>v>>k){
        s[]=;
        s[v-]=;
        for(i=(v-)/,j=;j*+<=k;i--,j++){
            s[i]=;
            s[v--i]=;
         }
         if(v<k){
            for(i=v;i<k;++i) s[i]=;
            v=k;
         }
         LL ans=;
         for(i=;i<v;++i){
            ans=(ans*+s[i])%mod;
         }
         cout<<ans<<endl;
     }
    return ;
 }

https://ac.nowcoder.com/acm/contest/272/B

考虑每一个点做出的贡献，通过这个点的路径数量是奇数就贡献一个A[u]，否则就是0。

求每个点的路径数量时，只要dfs出所有子树的节点数量即可。

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define pii pair<int,int>
 #define inf 0x3f3f3f3f
 #define pb push_back
 #define mp make_pair
 const LL mod=1e9+;
 const int maxn=;
 vector<int>g[maxn];
 LL A[maxn],n,ans=;
 int dfs(int u,int fa){
    int sum=,s=;
    for(int i=;i<g[u].size();++i){
        int v=g[u][i];
        if(v==fa) continue;
        int tmp=dfs(v,u);
        //s+=tmp;
        sum+=(tmp*(s+));
        s+=tmp;
        //if(u==1)cout<<"sum= "<<sum<<" "<<s<<endl;
        sum%=;
     }
     sum+=((LL)n-s-)*((LL)s+);
     sum%=;
     s++;
     if(sum)ans^=A[u];
     return s;
 }
 int main(){
    int i,j,u,v;
    scanf("%lld",&n);
    for(i=;i<n;++i){
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        g[v].push_back(u);
     }
     for(i=;i<=n;++i)scanf("%lld",A+i);
     dfs(,);
     printf("%lld\n",ans);
    return ;
 }

https://ac.nowcoder.com/acm/contest/272/C

观察发现n次操作之后白球个数的概率都是1/(n+1), 所以答案就是  1/(1+n) *(n+2)*(n+1)/2 = (n+2)/2

 #include<bits/stdc++.h>
 using namespace std;
 int main(){
     int n;
     scanf("%d",&n);
     printf("%.7f\n",(2.0+n)/);
     return ;
 }