## A - Buying Sweets

Topic link ：https://abc087.contest.atcoder.jp/tasks/abc087_a

Time limit : 2sec / Memory limit : 256MB

Score : `100` points

### Problem Statement

You went shopping to buy cakes and donuts with ` X` yen (the currency of Japan).

First, you bought one cake for ` A` yen at a cake shop. Then, you bought as many donuts as possible for

`yen each, at a donut shop.`

*B*How much do you have left after shopping?

### Constraints

`1≤`*A*,*B*≤1`000`*A*+*B*≤*X*≤10`000`,*X*and*A*are integers.*B*

### Input

Input is given from Standard Input in the following format:

X

A

B

### Output

Print the amount you have left after shopping.

### Sample Input 1

1234

150

100

### Sample Output 1

84

You have `1234−150=1084` yen left after buying a cake. With this amount, you can buy `10` donuts, after which you have `84` yen left.

### Sample Input 2

1000

108

108

### Sample Output 2

28

### Sample Input 3

579

123

456

### Sample Output 3

0

### Sample Input 4

7477

549

593

### Sample Output 4

405

#include<bits/stdc++.h>

using namespace std; int main()

{

int x,a,b;

while(cin>>x>>a>>b){

x-=a;

x%=b;

cout<<x<<endl;

}

return ;

}

## B - Coins

Topic link ：https://abc087.contest.atcoder.jp/tasks/abc087_b

Time limit : 2sec / Memory limit : 256MB

Score : `200` points

### Problem Statement

You have *A*`500`-yen coins, *B*`100`-yen coins and *C*`50`-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are ` X` yen in total?

Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

### Constraints

`0≤`*A*,*B*,*C*≤50*A*+*B*+*C*≥1`50≤`*X*≤20`000`,*A*and*B*are integers.*C*is a multiple of*X*`50`.

### Input

Input is given from Standard Input in the following format:

A

B

C

X

### Output

Print the number of ways to select coins.

### Sample Input 1

2

2

2

100

### Sample Output 1

2

There are two ways to satisfy the condition:

- Select zero
`500`-yen coins, one`100`-yen coin and zero`50`-yen coins. - Select zero
`500`-yen coins, zero`100`-yen coins and two`50`-yen coins.

### Sample Input 2

5

1

0

150

### Sample Output 2

0

Note that the total must be exactly ` X` yen.

### Sample Input 3

30

40

50

6000

### Sample Output 3

213

Answer key ： give change

#include<bits/stdc++.h>

using namespace std; int main()

{

int x,a,b,c;

while(cin>>a>>b>>c>>x){

x/=;

int sum=;

for(int i=;i<=a;i++){

for(int j=;j<=b;j++){

for(int k=;k<=c;k++){

if(i*+j*+k==x) sum++;

}

}

}

cout<<sum<<endl;

}

return ;

}

## C - Candies

Topic link ：https://abc087.contest.atcoder.jp/tasks/arc090_a

Time limit : 2sec / Memory limit : 256MB

Score : `300` points

### Problem Statement

We have a `2× N` grid. We will denote the square at the

`-th row and`

*i*`-th column (`

*j*`1≤`,

*i*≤2`1≤`) as

*j*≤*N*`(`.

*i*,*j*)You are initially in the top-left square, `(1,1)`. You will travel to the bottom-right square, `(2, N)`, by repeatedly moving right or down.

The square `( i,j)` contains

`candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.`

*A*_{i,j}At most how many candies can you collect when you choose the best way to travel?

### Constraints

`1≤`*N*≤100`1≤`(*A*_{i,j}≤100`1≤`,*i*≤2`1≤`)*j*≤*N*

### Input

Input is given from Standard Input in the following format:

N

A_{1,1}A_{1,2}…A_{1,N}

A_{2,1}A_{2,2}…A_{2,N}

### Output

Print the maximum number of candies that can be collected.

### Sample Input 1

5

3 2 2 4 1

1 2 2 2 1

### Sample Output 1

14

The number of collected candies will be maximized when you:

- move right three times, then move down once, then move right once.

### Sample Input 2

4

1 1 1 1

1 1 1 1

### Sample Output 2

5

You will always collect the same number of candies, regardless of how you travel.

### Sample Input 3

7

3 3 4 5 4 5 3

5 3 4 4 2 3 2

### Sample Output 3

29

### Sample Input 4

1

2

3

### Sample Output 4

5 Answer key ： Split into two arrays Use both prefix and Taking the maximum

#include<bits/stdc++.h>

using namespace std;

int a[],b[];

int suma[],sumb[];

int main()

{

int n;

while(cin>>n){

for(int i=;i<n;i++){

cin>>a[i];

}

for(int i=;i<n;i++){

cin>>b[i];

}

suma[]=a[];sumb[]=b[];

for(int i=;i<n;i++){

suma[i]=suma[i-]+a[i];

sumb[i]=sumb[i-]+b[i];

}

int maxn=suma[]+sumb[n-];

for(int i=;i<n;i++){

int sum=suma[i]+sumb[n-]-sumb[i-];

maxn=max(maxn,sum);

}

cout<<maxn<<endl;

}

return ;

}

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