Anton的数学与编程挑战-优快云博客

本文链接：https://blog.youkuaiyun.com/tianyuhang123/article/details/68070370

A - Anton and Polyhedrons

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:

$\text{[math]}$

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 s_i — the name of thei-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.

Example

Input

4
Icosahedron
Cube
Tetrahedron
Dodecahedron

Output

Input

3
Dodecahedron
Octahedron
Octahedron

Output

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.

水到境界的题，不讲了

ac代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main()
{
    string a;
    int x;
    while(cin>>x)
    {
        int sum=0;
        for(int i=0;i<x;i++)
        {
            cin>>a;
            if(a=="Tetrahedron")
            {
                sum+=4;
            }
            if(a=="Cube")
            {
                sum+=6;
            }
            if(a=="Octahedron")
            {
                sum+=8;
            }
            if(a=="Dodecahedron")
            {
                sum+=12;
            }
            if(a=="Icosahedron")
            {
                sum+=20;
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}

B - Anton and Classes

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 (l_1, i, r_1, i). Also he has m variants when he will attend programming classes, i-th variant is given by a period of time (l_2, i, r_2, 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 l_1, i and r_1, i (1 ≤ l_1, i ≤ r_1, 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 l_2, i and r_2, i (1 ≤ l_2, i ≤ r_2, 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.

Example

Input

Output

Input

Output

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.

需要干两件事，每件事给出多个时间段，让你选出两个时间段使得中间休息的时间最长

思路：分别求得每个人所有时间段中最先结束的时间和最后开始的时间，然后判断一下即可，如果两人存在不冲突的时间，那么肯定有一个人在前，一人在后，找一下最大值就可以了，所有时间都冲突就输出0

ac代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main()
{
int x,y;
int maxn1;
int maxn2;
int minn1;
int minn2;
int m,n;
while(cin>>x)
{
maxn1=0;
minn1=1000000000;
for(int i=0;i<x;i++)
{
scanf("%d%d",&m,&n);
if(m>maxn1)
maxn1=m;
if(n<minn1)
minn1=n;
}
cin>>y;
maxn2=0;
minn2=1000000000;
for(int i=0;i<y;i++)
{
scanf("%d%d",&m,&n);
if(m>maxn2)
maxn2=m;
if(n<minn2)
minn2=n;
}
int g=max(maxn1-minn2,maxn2-minn1);
if(g>=0)
cout<<g<<endl;
else
cout<<0<<endl;
}
return 0;
}

cf 785 水题