A. Anton and Polyhedrons
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:

  • Tetrahedron. Tetrahedron has 4 triangular faces.
  • Cube. Cube has 6 square faces.
  • Octahedron. Octahedron has 8 triangular faces.
  • Dodecahedron. Dodecahedron has 12 pentagonal faces.
  • Icosahedron. Icosahedron has 20 triangular faces.

All five kinds of polyhedrons are shown on the picture below:

Anton has a collection of n polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of polyhedrons in Anton's collection.

Each of the following n lines of the input contains a string si — the name of the i-th polyhedron in Anton's collection. The string can look like this:

  • "Tetrahedron" (without quotes), if the i-th polyhedron in Anton's collection is a tetrahedron.
  • "Cube" (without quotes), if the i-th polyhedron in Anton's collection is a cube.
  • "Octahedron" (without quotes), if the i-th polyhedron in Anton's collection is an octahedron.
  • "Dodecahedron" (without quotes), if the i-th polyhedron in Anton's collection is a dodecahedron.
  • "Icosahedron" (without quotes), if the i-th polyhedron in Anton's collection is an icosahedron.
Output

Output one number — the total number of faces in all the polyhedrons in Anton's collection.

Examples
Input
4
Icosahedron
Cube
Tetrahedron
Dodecahedron
Output
42
Input
3
Dodecahedron
Octahedron
Octahedron
Output
28
Note

In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.

题意：五种立方体 给你n个立方体 输出一共有多少个面

题解：水

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <stack>
 7 #include <queue>
 8 #include <cmath>
 9 #include <map>
10 #define ll  __int64
11 #define mod 1000000007
12 #define dazhi 2147483647
13 #define bug() printf("!!!!!!!")
14 #define M 100005
15 using namespace  std;
16 int n;
17 string str;
18 map<string ,int> mp;
19 int main()
20 {
21     mp["Tetrahedron"]=4;
22     mp["Cube"]=6;
23     mp["Octahedron"]=8;
24     mp["Dodecahedron"]=12;
25     mp["Icosahedron"]=20;
26     scanf("%d",&n);
27     int ans=0;
28     for(int i=1;i<=n;i++)
29     {
30         cin>>str;
31         ans+=mp[str];
32     }
33     cout<<ans<<endl;
34     return 0;
35 }

 

B. Anton and Classes
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Anton likes to play chess. Also he likes to do programming. No wonder that he decided to attend chess classes and programming classes.

Anton has n variants when he will attend chess classes, i-th variant is given by a period of time (l1, i, r1, i). Also he has m variants when he will attend programming classes, i-th variant is given by a period of time (l2, i, r2, i).

Anton needs to choose exactly one of n possible periods of time when he will attend chess classes and exactly one of m possible periods of time when he will attend programming classes. He wants to have a rest between classes, so from all the possible pairs of the periods he wants to choose the one where the distance between the periods is maximal.

The distance between periods (l1, r1) and (l2, r2) is the minimal possible distance between a point in the first period and a point in the second period, that is the minimal possible |i - j|, where l1 ≤ i ≤ r1 and l2 ≤ j ≤ r2. In particular, when the periods intersect, the distance between them is 0.

Anton wants to know how much time his rest between the classes will last in the best case. Help Anton and find this number!

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of time periods when Anton can attend chess classes.

Each of the following n lines of the input contains two integers l1, i and r1, i (1 ≤ l1, i ≤ r1, i ≤ 109) — the i-th variant of a period of time when Anton can attend chess classes.

The following line of the input contains a single integer m (1 ≤ m ≤ 200 000) — the number of time periods when Anton can attend programming classes.

Each of the following m lines of the input contains two integers l2, i and r2, i (1 ≤ l2, i ≤ r2, i ≤ 109) — the i-th variant of a period of time when Anton can attend programming classes.

Output

Output one integer — the maximal possible distance between time periods.

Examples
Input
3
1 5
2 6
2 3
2
2 4
6 8
Output
3
Input
3
1 5
2 6
3 7
2
2 4
1 4
Output
0
Note

In the first sample Anton can attend chess classes in the period (2, 3) and attend programming classes in the period (6, 8). It's not hard to see that in this case the distance between the periods will be equal to 3.

In the second sample if he chooses any pair of periods, they will intersect. So the answer is 0.

题意：给你n个区间 m个区间 在其中各选择一个区间 输出最大的区间间隔  两个选取的区间先后位置不确定

题解：水

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <stack>
 7 #include <queue>
 8 #include <cmath>
 9 #include <map>
10 #define ll  __int64
11 #define mod 1000000007
12 #define dazhi 2147483647
13 #define bug() printf("!!!!!!!")
14 #define M 100005
15 using namespace  std;
16 int n,m;
17 ll l,r;
18 int main()
19 {
20     scanf("%d",&n);
21     ll maxn1=0,minx1=1e9+10;
22     ll maxn2=0,minx2=1e9+10;
23     for(int i=1;i<=n;i++)
24     {
25         scanf("%I64d %I64d",&l,&r);
26         minx1=min(minx1,r);
27         maxn2=max(maxn2,l);
28     }
29     scanf("%d",&m);
30       for(int i=1;i<=m;i++)
31     {
32         scanf("%I64d %I64d",&l,&r);
33         maxn1=max(maxn1,l);
34         minx2=min(minx2,r);
35     }
36     if(maxn1<=minx1&&maxn2<=minx2)
37         cout<<"0"<<endl;
38     else
39         cout<<max(maxn1-minx1,maxn2-minx2)<<endl;
40     return 0;
41 }

 

C. Anton and Fairy Tale
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Anton likes to listen to fairy tales, especially when Danik, Anton's best friend, tells them. Right now Danik tells Anton a fairy tale:

"Once upon a time, there lived an emperor. He was very rich and had much grain. One day he ordered to build a huge barn to put there all his grain. Best builders were building that barn for three days and three nights. But they overlooked and there remained a little hole in the barn, from which every day sparrows came through. Here flew a sparrow, took a grain and flew away..."

More formally, the following takes place in the fairy tale. At the beginning of the first day the barn with the capacity of n grains was full. Then, every day (starting with the first day) the following happens:

  • m grains are brought to the barn. If m grains doesn't fit to the barn, the barn becomes full and the grains that doesn't fit are brought back (in this problem we can assume that the grains that doesn't fit to the barn are not taken into account).
  • Sparrows come and eat grain. In the i-th day i sparrows come, that is on the first day one sparrow come, on the second day two sparrows come and so on. Every sparrow eats one grain. If the barn is empty, a sparrow eats nothing.

Anton is tired of listening how Danik describes every sparrow that eats grain from the barn. Anton doesn't know when the fairy tale ends, so he asked you to determine, by the end of which day the barn will become empty for the first time. Help Anton and write a program that will determine the number of that day!

Input

The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1018) — the capacity of the barn and the number of grains that are brought every day.

Output

Output one integer — the number of the day when the barn will become empty for the first time. Days are numbered starting with one.

Examples
Input
5 2
Output
4
Input
8 1
Output
5
Note

In the first sample the capacity of the barn is five grains and two grains are brought every day. The following happens:

  • At the beginning of the first day grain is brought to the barn. It's full, so nothing happens.
  • At the end of the first day one sparrow comes and eats one grain, so 5 - 1 = 4 grains remain.
  • At the beginning of the second day two grains are brought. The barn becomes full and one grain doesn't fit to it.
  • At the end of the second day two sparrows come. 5 - 2 = 3 grains remain.
  • At the beginning of the third day two grains are brought. The barn becomes full again.
  • At the end of the third day three sparrows come and eat grain. 5 - 3 = 2 grains remain.
  • At the beginning of the fourth day grain is brought again. 2 + 2 = 4 grains remain.
  • At the end of the fourth day four sparrows come and eat grain. 4 - 4 = 0 grains remain. The barn is empty.

So the answer is 4, because by the end of the fourth day the barn becomes empty.

题意：容量为n的粮仓 从第i天减少i  每天最多填充m到粮仓 问第几天仓库的粮食储备为零

题解：二分答案

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <stack>
 7 #include <queue>
 8 #include <cmath>
 9 #include <map>
10 #define ll  __int64
11 #define mod 1000000007
12 #define dazhi 2147483647
13 #define bug() printf("!!!!!!!")
14 #define M 100005
15 using namespace  std;
16 ll n,m;
17 bool fun(ll t)
18 {
19     if((t-(m))>=2000000000)
20         return false;
21     ll exm=n-(1+t-(m+1))*(t-(m+1))/2;
22     if(exm>t)
23         return true;
24     else
25         return false;
26 }
27 int main()
28 {
29     scanf("%I64d %I64d",&n,&m);
30     if(n<=m){
31         printf("%I64d\n",n);
32      return 0;
33      }
34      ll l=m+1,r=1e18,mid;
35      while(l<r)
36      {
37          mid=(l+r)>>1;
38          if(fun(mid)){
39             l=mid+1;
40          }
41          else
42              r=mid;
43      }
44      cout<<l<<endl;
45     return 0;
46 }