Atcoder 092

A - Traveling Budget

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Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and Dyen if you buy an unlimited ticket.

Find the minimum total fare when the optimal choices are made for trains and buses.

Constraints

• 1≤A≤1 000
• 1≤B≤1 000
• 1≤C≤1 000
• 1≤D≤1 000
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

A
B
C
D

Output

Print the minimum total fare.

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Sample Input 1

Copy

600
300
220
420

Sample Output 1

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520

The train fare will be 600 yen if you buy ordinary tickets, and 300 yen if you buy an unlimited ticket. Thus, the optimal choice for trains is to buy an unlimited ticket for 300 yen. On the other hand, the optimal choice for buses is to buy ordinary tickets for 220 yen.

Therefore, the minimum total fare is 300+220=520 yen.

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Sample Input 2

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555
555
400
200

Sample Output 2

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755

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Sample Input 3

Copy

549
817
715
603

Sample Output 3

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1152

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 #include <string>
 5 #include <string.h>
 6 #include <cstring>
 7 #include <stdio.h>
 8 #define lowbit(x) x&(-x)
 9 
10 using namespace std;
11 const int maxn=105;
12 typedef long long ll;
13 
14 
15 int main(){
16     ios::sync_with_stdio(0);
17     ll a,b,c,d;
18     cin>>a>>b>>c>>d;
19     ll minn=min(a,b);
20     ll minn2=min(c,d);
21     cout<<minn+minn2<<endl;
22    return 0;
23 }

B - Chocolate

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Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1≤i≤N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (Ai+1)-th day, the (2Ai+1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces.

Find the number of chocolate pieces prepared at the beginning of the camp.

Constraints

• 1≤N≤100
• 1≤D≤100
• 1≤X≤100
• 1≤Ai≤100 (1≤i≤N)
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

N
D X
A1
A2
:
AN

Output

Find the number of chocolate pieces prepared at the beginning of the camp.

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Sample Input 1

Copy

3
7 1
2
5
10

Sample Output 1

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8

The camp has 3 participants and lasts for 7 days. Each participant eats chocolate pieces as follows:

• The first participant eats one chocolate piece on Day 1, 3, 5 and 7, for a total of four.
• The second participant eats one chocolate piece on Day 1 and 6, for a total of two.
• The third participant eats one chocolate piece only on Day 1, for a total of one.

Since the number of pieces remaining at the end of the camp is one, the number of pieces prepared at the beginning of the camp is 1+4+2+1=8.

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Sample Input 2

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2
8 20
1
10

Sample Output 2

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29

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Sample Input 3

Copy

5
30 44
26
18
81
18
6

Sample Output 3

Copy

56

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 #include <string>
 5 #include <string.h>
 6 #include <cstring>
 7 #include <stdio.h>
 8 #define lowbit(x) x&(-x)
 9 
10 using namespace std;
11 const int maxn=110;
12 typedef long long ll;
13 int a[maxn];
14 
15 int main(){
16     ios::sync_with_stdio(0);
17     int n,d,x;
18     cin>>n>>d>>x;
19     ll  m=0;
20     memset(a,0,sizeof(a));
21     for(int i=0;i<n;i++){
22         cin>>a[i];
23     }
24     int tmp=0;
25     for(int i=0;i<n;i++){
26         tmp=a[i];
27         if(tmp<d) {
28             tmp+=1;
29             m+=2;
30         }
31         else {
32                 ++m;
33                 continue;
34             }
35         while(tmp<d){
36              tmp+=a[i];
37              if(tmp<=d)
38                ++m;
39              else break;
40         }
41     }
42     cout<<m+x<<endl;
43    return 0;
44 }