A - Infinite Coins

题目链接:https://abc088.contest.atcoder.jp/tasks/abc088_a

Time limit : 2sec / Memory limit : 256MB

Score: 100 points

Problem Statement

E869120 has A 1-yen coins and infinitely many 500-yen coins.
Determine if he can pay exactly N yen using only these coins.

Constraints

  • N is an integer between 1 and 10000 (inclusive).
  • A is an integer between 0 and 1000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
A

Output

If E869120 can pay exactly N yen using only his 1-yen and 500-yen coins, print Yes; otherwise, print No.


Sample Input 1

Copy
2018
218

Sample Output 1

Copy
Yes

We can pay 2018 yen with four 500-yen coins and 18 1-yen coins, so the answer is Yes.


Sample Input 2

Copy
2763
0

Sample Output 2

Copy
No

When we have no 1-yen coins, we can only pay a multiple of 500 yen using only 500-yen coins. Since 2763 is not a multiple of 500, we cannot pay this amount.


Sample Input 3

Copy
37
514

Sample Output 3

Copy
Yes
     #include <iostream>
using namespace std;
int main()
{
int n,a;
while(cin>>n>>a){
int s=n%;
if(s<=a){
cout<<"Yes"<<endl;
}
else cout<<"No"<<endl;
}
return ;
}

B - Card Game for Two

题目链接:https://abc088.contest.atcoder.jp/tasks/abc088_b

Time limit : 2sec / Memory limit : 256MB

Score: 200 points

Problem Statement

We have N cards. A number ai is written on the i-th card.
Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first.
The game ends when all the cards are taken by the two players, and the
score of each player is the sum of the numbers written on the cards
he/she has taken. When both players take the optimal strategy to
maximize their scores, find Alice's score minus Bob's score.

Constraints

  • N is an integer between 1 and 100 (inclusive).
  • ai (1≤iN) is an integer between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

N
a1 a2 a3 aN

Output

Print Alice's score minus Bob's score when both players take the optimal strategy to maximize their scores.


Sample Input 1

Copy
2
3 1

Sample Output 1

Copy
2

First, Alice will take the card with 3. Then, Bob will take the card with 1. The difference of their scores will be 3 - 1 = 2.


Sample Input 2

Copy
3
2 7 4

Sample Output 2

Copy
5

First, Alice will take the card with 7. Then, Bob will take the card with 4. Lastly, Alice will take the card with 2. The difference of their scores will be 7 - 4 + 2 = 5. The difference of their scores will be 3 - 1 = 2.


Sample Input 3

Copy
4
20 18 2 18

Sample Output 3

Copy
18
     #include <iostream>
#include <algorithm>
using namespace std;
int a[]; bool cmp(int x,int y)
{
return x>y;
} int main()
{
int n;
while(cin>>n){
for(int i=;i<n;i++){
cin>>a[i];
}
sort(a,a+n,cmp);
int sum1=,sum2=;
for(int i=;i<n;i++){
if(i%==) sum1+=a[i];
else sum2+=a[i];
}
cout<<sum1-sum2<<endl;
}
return ;
}

C - Takahashi's Information

题目链接:https://abc088.contest.atcoder.jp/tasks/abc088_c

Time limit : 2sec / Memory limit : 256MB

Score: 300 points

Problem Statement

We have a 3×3 grid. A number ci,j is written in the square (i,j), where (i,j) denotes the square at the i-th row from the top and the j-th column from the left.
According to Takahashi, there are six integers a1,a2,a3,b1,b2,b3 whose values are fixed, and the number written in the square (i,j) is equal to ai+bj.
Determine if he is correct.

Constraints

  • ci,j (1≤i≤3,1≤j≤3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

c1,1 c1,2 c1,3
c2,1 c2,2 c2,3
c3,1 c3,2 c3,3

Output

If Takahashi's statement is correct, print Yes; otherwise, print No.


Sample Input 1

Copy
1 0 1
2 1 2
1 0 1

Sample Output 1

Copy
Yes

Takahashi is correct, since there are possible sets of integers such as: a1=0,a2=1,a3=0,b1=1,b2=0,b3=1.


Sample Input 2

Copy
2 2 2
2 1 2
2 2 2

Sample Output 2

Copy
No

Takahashi is incorrect in this case.


Sample Input 3

Copy
0 8 8
0 8 8
0 8 8

Sample Output 3

Copy
Yes

Sample Input 4

Copy
1 8 6
2 9 7
0 7 7

Sample Output 4

Copy
No

题解:求和然后去掉正对角线的元素 剩下的是正对角线元素的两倍
     #include <bits/stdc++.h>
using namespace std;
int a[][];
int main()
{
int sum=,sum1=;
for(int i=;i<;i++){
for(int j=;j<;j++){
cin>>a[i][j];
sum+=a[i][j];
if(i==j) sum1+=a[i][j];
}
}
if(sum-sum1==*sum1) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
return ;
}


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