Atcoder AGC 019 A,B

A - Ice Tea Store

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

You've come to your favorite store Infinitesco to buy some ice tea.

The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.

You want to buy exactly N liters of ice tea. How many yen do you have to spend?

Constraints

• 1≤Q,H,S,D≤10⁸
• 1≤N≤10⁹
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

Q H S D
N

Output

Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.

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Sample Input 1

20 30 70 90
3

Sample Output 1

150

Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90+30+30=150 yen.

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Sample Input 2

10000 1000 100 10
1

Sample Output 2

100

Even though a 2-liter bottle costs just 10 yen, you need only 1 liter. Thus, you have to buy a 1-liter bottle for 100 yen.

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Sample Input 3

10 100 1000 10000
1

Sample Output 3

40

Now it's better to buy four 0.25-liter bottles for 10+10+10+10=40 yen.

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Sample Input 4

12345678 87654321 12345678 87654321
123456789

Sample Output 4

1524157763907942

    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <queue>
    #include <stack>
    #include <cstdlib>
    #include <iomanip>
    #include <cmath>
    #include <cassert>
    #include <ctime>
    #include <map>
    #include <set>
    using namespace std;
    #define lowbit(x) (x&(-x))
    #define max(x,y) (x>=y?x:y)
    #define min(x,y) (x<=y?x:y)
    #define MAX 100000000000000000
    #define MOD 1000000007
    #define pi acos(-1.0)
    #define ei exp(1)
    #define PI 3.141592653589793238462
    #define ios() ios::sync_with_stdio(false)
    #define INF 1044266558
    #define mem(a) (memset(a,0,sizeof(a)))
    typedef long long ll;
    ll a[],n;
    int main()
    {
        while(scanf("%lld%lld%lld%lld",&a[],&a[],&a[],&a[])!=EOF)
        {
            scanf("%lld",&n);
            for(int i=;i<;i++) a[i]=min(a[i],*a[i-]);
            printf("%lld\n",n/*a[]+(n&)*a[]);
        }
        return ;
    }

B - Reverse and Compare

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

You have a string A=A₁A₂…A_n consisting of lowercase English letters.

You can choose any two indices i and j such that 1≤i≤j≤n and reverse substring A_iA_i+1…A_j.

You can perform this operation at most once.

How many different strings can you obtain?

Constraints

• 1≤|A|≤200,000
• A consists of lowercase English letters.

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Input

Input is given from Standard Input in the following format:

A

Output

Print the number of different strings you can obtain by reversing any substring in A at most once.

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Sample Input 1

aatt

Sample Output 1

5

You can obtain aatt (don't do anything), atat (reverse A[2..3]), atta (reverse A[2..4]), ttaa (reverse A[1..4]) and taat (reverse A[1..3]).

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Sample Input 2

xxxxxxxxxx

Sample Output 2

1

Whatever substring you reverse, you'll always get xxxxxxxxxx.

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Sample Input 3

abracadabra

Sample Output 3

44
预处理。

    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <queue>
    #include <stack>
    #include <cstdlib>
    #include <iomanip>
    #include <cmath>
    #include <cassert>
    #include <ctime>
    #include <map>
    #include <set>
    using namespace std;
    #define lowbit(x) (x&(-x))
    #define max(x,y) (x>=y?x:y)
    #define min(x,y) (x<=y?x:y)
    #define MAX 100000000000000000
    #define MOD 1000000007
    #define pi acos(-1.0)
    #define ei exp(1)
    #define PI 3.141592653589793238462
    #define ios() ios::sync_with_stdio(false)
    #define INF 1044266558
    #define mem(a) (memset(a,0,sizeof(a)))
    typedef long long ll;
    ll a[][],ans;
    char s[];
    int main()
    {
        while(scanf("%s",s+)!=EOF)
        {
            memset(a,,sizeof(a));
            int len=strlen(s+);
            ans=;
            for(int i=;i<=len;i++)
            {
                for(int j=;j<;j++)
                    a[i][j]=a[i-][j];
                a[i][s[i]-'a']++;
            }
            for(int i=;i<=len;i++)
                ans+=len-i-a[len][s[i]-'a']+a[i][s[i]-'a'];
            printf("%lld\n",ans+);
        }
        return ;
    }