Codeforces Round #308 (Div. 2)

A. Vanya and Table

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Vanya has a table consisting of 100 rows, each row contains 100 cells.
The rows are numbered by integers from 1 to 100 from
bottom to top, the columns are numbered from 1 to 100 from
left to right.

In this table, Vanya chose n rectangles with sides that go along borders of squares (some rectangles probably occur multiple times).
After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result.

Input

The first line contains integer n (1 ≤ n ≤ 100)
— the number of rectangles.

Each of the following n lines contains four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ 100, 1 ≤ y1 ≤ y2 ≤ 100),
where x1 and y1 are
the number of the column and row of the lower left cell and x2 and y2 are
the number of the column and row of the upper right cell of a rectangle.

Output

In a single line print the sum of all values in the cells of the table.

Sample test(s)

input

2
1 1 2 3
2 2 3 3

output

10

input

2
1 1 3 3
1 1 3 3

output

18

Note

Note to the first sample test:

Values of the table in the first three rows and columns will be as follows:

121
121
110So, the sum of values will be equal to 10.

Note to the second sample test:

Values of the table in the first three rows and columns will be as follows:

222

222

222

So, the sum of values will be equal to 18.

求一共几个数

#include <bits/stdc++.h>
using namespace std;
int n;
int x1,yy1,x2,y2;
int main()
{
cin>>n;
int sum=0;
while(n--)
{
cin>>x1>>yy1>>x2>>y2;
sum+=(x2-x1+1)*(y2-yy1+1);
}
cout<<sum;
return 0;
}

B. Vanya and Books

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should
be assigned with a number from 1 to n.
Naturally, distinct books should be assigned distinct numbers.

Vanya wants to know how many digits he will have to write down as he labels the books.

Input

The first line contains integer n (1 ≤ n ≤ 109)
— the number of books in the library.

Output

Print the number of digits needed to number all the books.

Sample test(s)

input

13

output

17

input

4

output

4

Note

Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.

Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.

求1-n的数一共有几位数字

#include <bits/stdc++.h>
using namespace std;
long long n;
int main()
{
cin>>n;
long long a=9,b=0;
long long sum=0;
int k=1;
while(n>=a)
{
sum+=(a-b)*k;
a*=10;a+=9;
b*=10;b+=9;
k++;
}
sum+=(n-b)*k;
cout<<sum;
return 0;
}

C. Vanya and Scales

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams
where w is some integer not less than 2(exactly
one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can
be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and
some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.

Input

The first line contains two integers w, m (2 ≤ w ≤ 109, 1 ≤ m ≤ 109)
— the number defining the masses of the weights and the mass of the item.

Output

Print word 'YES' if the item can be weighted and 'NO' if it
cannot.

Sample test(s)

input

3 7

output

YES

input

100 99

output

YES

input

100 50

output

NO

Note

Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3,
and the second pan can have two weights of masses 9 and 1,
correspondingly. Then 7 + 3 = 9 + 1.

Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1,
and the second pan can have the weight of mass 100.

Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.

砝码是w的0到100次方，求能否称出重量m

dfs随便搞一搞

左面加，右面加，不加

#include <bits/stdc++.h>
using namespace std;
long long w,m;
int flag;
void dfs(long long b,long long c)
{
if(c==m)
{
flag=1;
return ;
}
if(flag) return ;
if(abs(b)<=m*w)
{
dfs(b*w,c+b);
dfs(b*w,c-b);
dfs(b*w,c);
}
return ;
}
int main()
{
flag=0;
cin>>w>>m;
if(w<=3)
{
cout<<"YES"<<endl;
return 0;
}
dfs(1,0);
if(flag) cout<<"YES";
else cout<<"NO";
return 0;
}

D. Vanya and Triangles

time limit per test
4 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result
many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.

Input

The first line contains integer n (1 ≤ n ≤ 2000)
— the number of the points painted on the plane.

Next n lines contain two integers each xi, yi ( - 100 ≤ xi, yi ≤ 100)
— the coordinates of the i-th point. It is guaranteed that no two given points coincide.

Output

In the first line print an integer — the number of triangles with the non-zero area among the painted points.

Sample test(s)

input

40 01 12 02 2

output

3

input

30 01 12 0

output

1

input

11 1

output

0

Note

Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).

Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).

Note to the third sample test. A single point doesn't form a single triangle.

求n个点能构成多少个三角形。

先抛出重复点，然后枚举两个点，确定不在这条直线上的点的数目。在这条直线的点的数目可以通过，gcd找一下整点，再-100 - 100 找出是否这条线的点有没有点。。。

最后数/6

#include <bits/stdc++.h>
using namespace std;
#define Max 2333int x[2333];
int y[2333];
int gcd(int a,int b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
int mp[233][233]={};
int main()
{
int n;
cin>>n;
int aa,bb;
for(int i=1;i<=n;i++)
{
cin>>aa>>bb;
aa+=100;
bb+=100;
mp[aa][bb]=1;
}
int top=0;
for(int i=0;i<=200;i++)
{
for(int j=0;j<=200;j++)
{
if(mp[i][j])
{
x[++top]=i;
y[top]=j;
}
}
}
n=top;
long long sum=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i!=j)
{
int u=-1;
int xa=x[i],ya=y[i];
int xb=x[j],yb=y[j];
int xx=xa-xb;
int yy=ya-yb;
int g;
if(xx==0||yy==0)
{
if(xx==0)
{
yy=1;
}
else xx=1;
}
else {g=gcd(xx,yy);
xx/=g;
yy/=g;}
int tx=xa;
int ty=ya;
while(tx<=200&&ty<=200&&tx>=0&&ty>=0)
{
tx+=xx;
ty+=yy;
if(!(tx<=200&&ty<=200&&tx>=0&&ty>=0)) break;
if(mp[tx][ty]) u++;
}
tx=xa;
ty=ya;
while(tx<=200&&ty<=200&&tx>=0&&ty>=0)
{
tx-=xx;
ty-=yy;
if(!(tx<=200&&ty<=200&&tx>=0&&ty>=0)) break;
if(mp[tx][ty]) u++;
}
sum+=(n-2-u);
}
}
}
cout<<sum/6;
return 0;
}

E. Vanya and Brackets

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vanya is doing his maths homework. He has an expression of form

,
where x1, x2, ..., xn are
digits from 1 to 9, and sign

represents
either a plus '+' or the multiplication sign '*'. Vanya needs
to add one pair of brackets in this expression so that to maximize the value of the resulting expression.

Input

The first line contains expression s (1 ≤ |s| ≤ 5001, |s| is
odd), its odd positions only contain digits from 1 to 9,
and even positions only contain signs  +  and  * .

The number of signs  *  doesn't exceed 15.

Output

In the first line print the maximum possible value of an expression.

Sample test(s)

input

3+5*7+8*4

output

303

input

2+3*5

output

25

input

3*4*5

output

60

Note

Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.

Note to the second sample test. (2 + 3) * 5 = 25.

Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).

我们要不要祭出python大法呢

s = raw_input()
ans = eval(s)
l = len(s)
a = []
for i in range (l):
if (s[i] == '*'):
a.append(i)
for i in a:
s1 = "(" + s[:i] + ")" + s[i:]
ans = max(ans, eval(s1))

for j in a:
if j > i:
if s[j] == '*':
s1 = s[:i+1] + "(" + s[i+1:j] + ")"+s[j:]
ans = max(ans, eval(s1))
s1 = s[:i+1] + "(" + s[i + 1:] + ")"
ans = max(ans, eval(s1))

print ans