Codeforces Round #355 (Div. 2)
A
弯腰
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<queue> 5 #include<stack> 6 #include<algorithm> 7 using namespace std; 8 #define clc(a,b) memset(a,b,sizeof(a)) 9 #define inf 0x3f3f3f3f 10 const int N=10010; 11 #define LL long long 12 // inline int r(){ 13 // int x=0,f=1;char ch=getchar(); 14 // while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();} 15 // while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 16 // return x*f; 17 // } 18 19 int main(){ 20 int n,m; 21 int ans=0; 22 scanf("%d%d",&n,&m); 23 for(int i=0;i<n;i++){ 24 int x; 25 cin>>x; 26 if(x<=m) 27 ans+=1; 28 else 29 ans+=2; 30 } 31 cout<<ans<<endl; 32 return 0; 33 }
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
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<queue> 5 #include<stack> 6 #include<algorithm> 7 using namespace std; 8 #define clc(a,b) memset(a,b,sizeof(a)) 9 #define inf 0x3f3f3f3f 10 const int N=10010; 11 #define LL long long 12 void fre(){freopen("in.txt","r",stdin); } 13 // inline int r(){ 14 // int x=0,f=1;char ch=getchar(); 15 // while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();} 16 // while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 17 // return x*f; 18 // } 19 int a[100010]; 20 21 int main(){ 22 // fre(); 23 int n,h,k; 24 cin>>n>>h>>k; 25 for(int i=1;i<=n;i++){ 26 scanf("%d",&a[i]); 27 } 28 LL sum=0; 29 LL ans=0; 30 for(int i=1;i<=n;i++){ 31 sum+=a[i]; 32 while(sum+a[i+1]<=h){ 33 sum+=a[i+1]; 34 i++; 35 if(i>n) 36 break; 37 } 38 // cout<<i<<" "<<sum<<endl; 39 while(1){ 40 int tem=sum; 41 sum%=k; 42 ans+=tem/k; 43 if(sum<=0) 44 { 45 sum=0; 46 break; 47 } 48 if(i<n){ 49 if(sum+a[i+1]<=h) 50 break; 51 } 52 if(sum<k&&sum>0){ 53 sum=0; 54 ans++; 55 // cout<<sum<<" "<<ans<<endl; 56 break; 57 } 58 } 59 // cout<<i<<" "<<sum<<endl; 60 } 61 cout<<ans<<endl; 62 return 0; 63 }
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
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<queue> 5 #include<stack> 6 #include<algorithm> 7 using namespace std; 8 #define clc(a,b) memset(a,b,sizeof(a)) 9 #define inf 0x3f3f3f3f 10 const int N=10010; 11 const int MOD = 1e9+7; 12 #define LL long long 13 void fre(){freopen("in.txt","r",stdin); } 14 // inline int r(){ 15 // int x=0,f=1;char ch=getchar(); 16 // while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();} 17 // while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 18 // return x*f; 19 // } 20 LL pow_m(LL a,LL b) 21 { 22 LL ans = 1; 23 a %= MOD; 24 while(b) 25 { 26 if(b & 1) 27 { 28 ans = ans * a % MOD; 29 b--; 30 } 31 b >>= 1; 32 a = a * a % MOD; 33 } 34 return ans; 35 } 36 37 int fun(int x){ 38 int num=0; 39 int cnt=6; 40 while(cnt){ 41 if((x&1)==0) 42 num++; 43 x>>=1; 44 cnt--; 45 } 46 return num; 47 } 48 string s; 49 bool vis[100]; 50 int main(){ 51 // freopen("in.txt","r",stdin); 52 getline(cin,s); 53 LL num=0; 54 LL ans=0; 55 clc(vis,false); 56 for(int i=0;i<s.length();i++){ 57 int x; 58 if(s[i]>='0'&&s[i]<='9'){ 59 x=s[i]-'0'; 60 num+=fun(x); 61 } 62 else if(s[i]>='A'&&s[i]<='Z'){ 63 x=s[i]-'A'+10; 64 num+=fun(x); 65 } 66 else if(s[i]>='a'&&s[i]<='z'){ 67 x=s[i]-'a'+36; 68 num+=fun(x); 69 } 70 else if(s[i]=='-'){ 71 num+=fun(62); 72 } 73 else{ 74 num+=fun(63); 75 } 76 } 77 // cout<<num<<endl; 78 ans=pow_m(3,num); 79 printf("%I64d\n",ans); 80 return 0; 81 }
