校个人赛——02

A - Infinite Sequence
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

Description

Vasya likes everything infinite. Now he is studying the properties of a sequence s, such that its first element is equal to a (s1 = a),
and the difference between any two neighbouring elements is equal to c (si - si - 1 = c).
In particular, Vasya wonders if his favourite integerb appears in this sequence, that is, there exists a positive integer i, such
that si = b. Of course, you are the person he asks for a help.

Input

The first line of the input contain three integers a, b and c ( - 109 ≤ a, b, c ≤ 109) —
the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.

Output

If b appears in the sequence s print "YES" (without quotes), otherwise print "NO"
(without quotes).

Sample Input

Input

1 7 3

Output

YES

Input

10 10 0

Output

YES

Input

1 -4 5

Output

NO

Input

0 60 50

Output

NO

分类讨论

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<cmath>
#include<algorithm>
#define INF (1ll<<60)-1
using namespace std;
int my_max(int x,int y) {return x>y?x:y;}
int my_min(int x,int y) {return x>y?y:x;}

int main()
{
long a,b,c,d,e;
while(scanf("%ld %ld %ld",&a,&b,&c)!=EOF)
{
if(b>a)
{
if(c<=0)
{
printf("NO\n");
}
else
{
e=(b-a)%c;
if(e==0) printf("YES\n");
else printf("NO\n");
}
}
else if(b<a)
{
if(c>=0)
{
printf("NO\n");
}
else
{
e=(b-a)%(-1*c);
if(e==0) printf("YES\n");
else printf("NO\n");
}

}
else
{
printf("YES\n");
}
}
return 0;
}

注意事项：除数不能为0

B - Restoring Painting
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

Description

Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.

The painting is a square 3 × 3, each cell contains a single integer from 1 to n,
and different cells may contain either different or equal integers.
The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
Four elements a, b, c and d are
known and are located as shown on the picture below.



Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.

Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.

Input

The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) —
maximum possible value of an integer in the cell and four integers that Vasya remembers.

Output

Print one integer — the number of distinct valid squares.

Sample Input

Input

2 1 1 1 2

Output

2

Input

3 3 1 2 3

Output

6

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<cmath>
#include<algorithm>
#define INF (1ll<<60)-1
using namespace std;
int my_max(int x,int y) {return x>y?x:y;}
int my_min(int x,int y) {return x>y?y:x;}

int main()
{
int n,a,b,c,d,j,count;
while(scanf("%d %d %d %d %d",&n,&a,&b,&c,&d)!=EOF)
{
count=0;
for(j=1;j<=n;j++)
{
sum=a+b+j;
if(sum-a-c>n||sum-a-c<=0) continue;
if(sum-d-c>n||sum-d-c<=0) continue;
if(sum-b-d>n||sum-b-d<=0) continue;
count++ ;
}
printf("%I64d\n",(long long)count*(long long)n);
}
return 0;
}

简单的数学小题：
注意其输出形式！！！

为long long 形式，而不是int