Codeforces Round #355 (Div. 2)

A

弯腰

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<queue>
 #include<stack>
 #include<algorithm>
 using namespace std;
 #define clc(a,b) memset(a,b,sizeof(a))
 #define inf 0x3f3f3f3f
 const int N=;
 #define LL long long
 // inline int r(){
 //     int x=0,f=1;char ch=getchar();
 //     while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();}
 //     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 //     return x*f;
 // }

 int main(){
     int n,m;
     int ans=;
     scanf("%d%d",&n,&m);
     for(int i=;i<n;i++){
        int x;
        cin>>x;
        if(x<=m)
         ans+=;
        else
         ans+=;
     }
     cout<<ans<<endl;
     return ;
 }

B

处理器一秒能处理k个东西，缓冲区最多不能超过h个，问几秒处理完

模拟，注意h很大，k很小的情况！！所以必须用除法避免超时

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<queue>
 #include<stack>
 #include<algorithm>
 using namespace std;
 #define clc(a,b) memset(a,b,sizeof(a))
 #define inf 0x3f3f3f3f
 const int N=;
 #define LL long long
 void fre(){freopen("in.txt","r",stdin); }
 // inline int r(){
 //     int x=0,f=1;char ch=getchar();
 //     while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();}
 //     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 //     return x*f;
 // }
 int a[];

 int main(){
     // fre();
     int n,h,k;
     cin>>n>>h>>k;
     for(int i=;i<=n;i++){
        scanf("%d",&a[i]);
     }
     LL sum=;
     LL ans=;
     for(int i=;i<=n;i++){
        sum+=a[i];
        while(sum+a[i+]<=h){
           sum+=a[i+];
           i++;
           if(i>n)
             break;
        }
        // cout<<i<<" "<<sum<<endl;
        while(){
           int tem=sum;
           sum%=k;
           ans+=tem/k;
           if(sum<=)
           {
             sum=;
             break;
           }
           if(i<n){
             if(sum+a[i+]<=h)
                break;
           }
           if(sum<k&&sum>){
              sum=;
              ans++;
              // cout<<sum<<" "<<ans<<endl;
              break;
           }
        }
        // cout<<i<<" "<<sum<<endl;
     }
     cout<<ans<<endl;
     return ;
 }

C

统计每个字符对应的数字有几个0，答案就是pow（3，n）。二进制是六位

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<queue>
 #include<stack>
 #include<algorithm>
 using namespace std;
 #define clc(a,b) memset(a,b,sizeof(a))
 #define inf 0x3f3f3f3f
 const int N=;
 const int MOD = 1e9+;
 #define LL long long
 void fre(){freopen("in.txt","r",stdin); }
 // inline int r(){
 //     int x=0,f=1;char ch=getchar();
 //     while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();}
 //     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 //     return x*f;
 // }
 LL pow_m(LL a,LL b)
 {
     LL ans = ;
     a %= MOD;
 while(b)
 {
         if(b & )
         {
             ans = ans * a % MOD;
             b--;
         }
         b >>= ;
         a = a * a % MOD;
     }
     return ans;
 }  

 int fun(int x){
     int num=;
     int cnt=;
     while(cnt){
        if((x&)==)
           num++;
        x>>=;
        cnt--;
     }
     return num;
 }
 string s;
 bool vis[];
 int main(){
     // freopen("in.txt","r",stdin);
     getline(cin,s);
     LL num=;
     LL ans=;
     clc(vis,false);
     for(int i=;i<s.length();i++){
         int x;
         if(s[i]>=''&&s[i]<=''){
            x=s[i]-'';
            num+=fun(x);
         }
         else if(s[i]>='A'&&s[i]<='Z'){
            x=s[i]-'A'+;
            num+=fun(x);
         }
         else if(s[i]>='a'&&s[i]<='z'){
             x=s[i]-'a'+;
             num+=fun(x);
         }
         else if(s[i]=='-'){
            num+=fun();
         }
         else{
            num+=fun();
         }
     }
     // cout<<num<<endl;
     ans=pow_m(,num);
     printf("%I64d\n",ans);
     return ;
 }