Codeforces Round #239 (Div. 二) A~D

Codeforces Round #239 (Div. 2) A~D

Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.

There are n cashiers at the exit from the supermarket. At the moment the queue for the i-th cashier already has ki people. The j-th person standing in the queue to the i-th cashier has mi, j items in the basket. Vasya knows that:

the cashier needs 5 seconds to scan one item;
after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.

Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of cashes in the shop. The second line contains n space-separated integers: k1, k2, ..., kn (1 ≤ ki ≤ 100), where ki is the number of people in the queue to the i-th cashier.

The i-th of the next n lines contains ki space-separated integers: mi, 1, mi, 2, ..., mi, ki (1 ≤ mi, j ≤ 100) — the number of products thej-th person in the queue for the i-th cash has.

Output

Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.

Sample test(s)

input
1
1
1

output
20

input
4
1 4 3 2
100
1 2 2 3
1 9 1
7 8

output
100

Note

In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100·5 + 15 = 515 seconds. But if he chooses the second queue, he will need 1·5 + 2·5 + 2·5 + 3·5 + 4·15 = 100 seconds. He will need 1·5 + 9·5 + 1·5 + 3·15 = 100 seconds for the third one and 7·5 + 8·5 + 2·15 = 105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int INF=0x3f3f3f3f;
int n,k[110],m[110][110];

int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>k[i];
    int ans=INF;
    for(int i=0;i<n;i++)
    {
        int temp=k[i]*15;;
        for(int j=0;j<k[i];j++)
        {
            cin>>m[i][j];
            temp+=m[i][j]*5;
        }
        ans=min(ans,temp);
    }
    cout<<ans<<endl;
    return 0;
}

Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.

The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.

Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get.

Input

The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.

The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make.

Output

Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer.

Sample test(s)

input
aaabbac
aabbccac

output
6

input
a
z

output
-1

Note

In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color aand cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.

In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

char sa[1100],sb[1100];

int color1[30],color2[30];

int main()
{
    cin>>sa>>sb;
    int n=strlen(sa),m=strlen(sb);
    for(int i=0;i<n;i++)
    {
        color1[sa[i]-'a']++;
    }
    for(int i=0;i<m;i++)
    {
        color2[sb[i]-'a']++;
    }
    int ans=0,flag=true;
    for(int i=0;i<30;i++)
    {
        ans+=min(color1[i],color2[i]);
        if(color2[i]&&color1[i]==0)
            flag=false;
    }
    if(ans==0||flag==false) puts("-1");
    else cout<<ans<<endl;
    return 0;
}

There is a right triangle with legs of length a and b. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.

Input

The first line contains two integers a, b (1 ≤ a, b ≤ 1000), separated by a single space.

Output

In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.

Sample test(s)

input
1 1

output
NO

input
5 5

output
YES
2 1
5 5
-2 4

input
5 10

output
YES
-10 4
-2 -2
1 2

可能组合太少，直接暴力枚举

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>

using namespace std;

int a,b;
typedef pair<int,int> pII;

set<pII> sa,sb;

int main()
{
    cin>>a>>b;
    for(int i=0;i<1000;i++)
    {
        if(i*i>a*a) break;
        int temp=a*a-i*i;
        int j=(int)(sqrt(temp));
        if(j*j==temp)
        {
            if(i==0||j==0) continue;
            sa.insert(make_pair(i,j));
            sa.insert(make_pair(i,-j));
            sa.insert(make_pair(-i,j));
            sa.insert(make_pair(-i,-j));
        }
    }
    for(int i=0;i<1000;i++)
    {
        if(i*i>b*b) break;
        int temp=b*b-i*i;
        int j=(int)(sqrt(temp));
        if(j*j==temp)
        {
            if(i==0||j==0) continue;
            sb.insert(make_pair(i,j));
            sb.insert(make_pair(i,-j));
            sb.insert(make_pair(-i,j));
            sb.insert(make_pair(-i,-j));
        }
    }

    set<pII>::iterator ita,itb;
    for(ita=sa.begin();ita!=sa.end();ita++)
    {
        int xa,xb,ya,yb;
        xa=ita->first,ya=ita->second;
        for(itb=sb.begin();itb!=sb.end();itb++)
        {
            xb=itb->first,yb=itb->second;
            if(xa*xb+ya*yb==0)
            {
                if(xa==0||xb==0||ya==0||yb==0||xa==xb||ya==yb) continue;
                else
                {
                    puts("YES");
                    cout<<0<<" "<<0<<endl;
                    cout<<xa<<" "<<ya<<endl;
                    cout<<xb<<" "<<yb<<endl;
                    return 0;
                }
            }
        }
    }
    puts("NO");
    return 0;
}

One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.

The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room numberpi, where 1 ≤ pi ≤ i.

In order not to get lost, Vasya decided to act as follows.

Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.

Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.

Input

The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pidenotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.

Output

Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).

Sample test(s)

input
2
1 2

output
4

input
4
1 1 2 3

output
20

input
5
1 1 1 1 1

output
62

递推，经过前面一个点两次才能走到下一个点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const long long int MOD=1e9+7;

long long int dp[1100];
int n,p[1100];

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>p[i];
    for(int i=2;i<=n+1;i++)
    {
        dp[i]=(2*dp[i-1]-dp[p[i-1]]+2+MOD)%MOD;
    }
    cout<<dp[n+1]%MOD<<endl;
    return 0;
}

Codeforces Round #239 (Div. 二) A~D

相关推荐