NOIP2017模拟赛 senior 6.29 T2 小T的钢琴（piano）

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<cmath>
  6 #include<string>
  7 #define INT_MAX_ 0x7fffffff
  8 #define LONG_MAX_ 0x7fffffffffffffffll
  9 #define pi acos(-1)
 10 #define N 100010
 11 #define M 200010
 12 using namespace std;
 13 
 14 inline int read()
 15 {
 16     int data=0,w=1;
 17     char ch=0;
 18     while(ch!=‘-‘ && (ch<‘0‘ || ch>‘9‘)) ch=getchar();
 19     if(ch==‘-‘) w=-1,ch=getchar();
 20     while(ch>=‘0‘ && ch<=‘9‘) data=data*10+ch-‘0‘,ch=getchar();
 21     return data*w;
 22 }
 23 
 24 inline void write(int x)
 25 {
 26     if(x<0) putchar(‘-‘),x=-x;
 27     if(x>9) write(x/10);
 28     putchar(x%10+‘0‘);
 29 }
 30 
 31 struct Data
 32 {
 33     int value,number;
 34 } data[N];
 35 inline bool my_comp(const Data &a, const Data &b)
 36 {
 37     return a.value < b.value;
 38 }
 39 
 40 int b[M],ans[N];
 41 char Little[25];
 42 char str[25];
 43 int n,r;
 44 
 45 int main()
 46 {
 47     freopen("piano.in","r",stdin);
 48     freopen("piano.out","w",stdout);
 49 
 50     n = read();
 51     r = read();
 52     for(int i = 1; i <= n; i++)
 53     {
 54         scanf("%s",Little);
 55         int lenth = strlen(Little);
 56         data[i].number = i;
 57         data[data[i].number].value = (Little[0] - ‘A‘ + 1) * 10 + Little[1] - ‘0‘;
 58         if(lenth > 2 && Little[2] == ‘.‘)
 59         {
 60             for(int j = 3; j < lenth; j++)
 61             {
 62                 if(Little[j] == ‘(‘ || Little[j] == ‘[‘ || Little[j] == ‘{‘) break;
 63                 data[i].value = data[i].value * 10 + Little[j] - ‘0‘;
 64             }
 65         }
 66     }
 67     sort(data+1,data+1+n,my_comp);
 68     while(r--)
 69     {
 70         int li = read();
 71         for(int i = 1; i <= li; i++)
 72         {
 73             scanf("%s",str);
 74             int lenth = strlen(str);
 75             b[i] = (str[0] - ‘A‘ + 1) * 10 + str[1] - ‘0‘;
 76             if(lenth > 2 && str[2] == ‘.‘)
 77             {
 78                 for(int j = 3; j < lenth; j++)
 79                 {
 80                     if(str[j] == ‘(‘ || str[j] == ‘[‘ || str[j] == ‘{‘) break;
 81                     b[i] = b[i] * 10 + str[j] - ‘0‘;
 82                 }
 83             }
 84         }
 85         ans[0] = 0;
 86         for(int i = 1; i <= li; i++)
 87         {
 88             int left = 1, right = n;
 89             while(left < right)
 90             {
 91                 int mid = (left + right) / 2;
 92                 if(data[mid].value >= b[i])
 93                 {
 94                     right = mid;
 95                 }
 96                 else
 97                 {
 98                     left = mid + 1;
 99                 }
100             }
101             if(data[left].value != b[i]) continue;
102             if(ans[ans[0]] < data[left].number) ans[++ans[0]] = data[left].number;
103             else
104             {
105                 int aider = lower_bound(ans+1,ans+1+ans[0],data[left].number) - ans;
106                 ans[aider] = data[left].number;
107             }
108         }
109         printf("%.6lf\n",ans[0] * 1.0 / li);
110     }
111 
112     return 0;
113 }