Educational Codeforces Round 35 (Rated for Div. 2)A，B，C

A. Nearest Minimums

time limit per test
 2 seconds

memory limit per test
 256 megabytes

input
 standard input

output
 standard output

You are given an array of n integer numbers a0, a1, ..., an - 1.
Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.

Input

The first line contains positive integer n (2 ≤ n ≤ 105)
— size of the given array. The second line contains n integers a0, a1, ..., an - 1(1 ≤ ai ≤ 109)
— elements of the array. It is guaranteed that in the array a minimum occurs at least two times.

Output

Print the only number — distance between two nearest minimums in the array.

Examples

input

2
3 3

output

1

input

3
5 6 5

output

2

input

9
2 1 3 5 4 1 2 3 1

output

3

题解：用结构体记录输入时的下标，然后排序，直接暴力最小值部分就行了，记录下标的最大差值

代码

#include<cstdio>
#include<algorithm>
#include<math.h>
#define M 100010
#define INF 0x3f3f3f3f

using namespace std;

struct node
{
int x;
int y;
}a[M];

bool cmp(node A, node B)
{
if(A.x != B.x)return A.x < B.x;
else return A.y < B.y;
}

int main()
{
int n;
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%d", &a[i].x);
a[i].y = i;
}
sort(a + 1, a + n + 1, cmp);
/*for(int i = 1; i <= n; i++)
{
printf("%d %d\n", a[i].x, a[i].y);
}*/
int minn = INF;
int pos;
for(int i = 1; i <= n; i++)
if(a[i].x == a[1].x)pos = i;
//printf("pos = %d\n", pos);
for(int i = 2; i <= pos; i++)
{
minn = min(minn, (int)fabs(a[i].y - a[i - 1].y));
}

printf("%d\n", minn);
return 0;
}


B. Two Cakes

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into a pieces,
and the second one — into b pieces.

Ivan knows that there will be n people at the celebration (including himself), so Ivan has set n plates
for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:
Each piece of each cake is put on some plate;
Each plate contains at least one piece of cake;
No plate contains pieces of both cakes.

To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number x such
that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least x pieces of cake.

Help Ivan to calculate this number x!

Input

The first line contains three integers n, a and b (1 ≤ a, b ≤ 100, 2 ≤ n ≤ a + b)
— the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.

Output

Print the maximum possible number x such that Ivan can distribute the cake in such a way that each plate will contain at
least x pieces of cake.

Examples

input

5 2 3

output

1

input

4 7 10

output

3

Note

In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.

In the second example you can have two plates with 3 and 4 pieces
of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.

题解：最小值最大化问题
方法一：看最后一个要求，同一个盘子里不能有两种蛋糕，说明盘子是被分成两个部分的，一份放a，一份放b，所以直接暴力盘中数就好了。
代码

#include<cstdio>
#include<algorithm>

using namespace std;

int main()
{
int n, a, b;
while(scanf("%d %d %d", &n, &a, &b) != EOF)
{
if(n == a + b)
{
printf("1\n");
continue;
}
int x, y;
int minn, maxx = 0;
for(int i = 1; i <= n - 1; i++)
{
x = a / i;
y = b / (n - i);
minn = min(x, y);
maxx = max(maxx, minn);
}
printf("%d\n", maxx);
}
return 0;
}

方法二：二分法
代码

#include<cstdio>

int n, a, b;

bool judge(int x)
{
if(a / x + b / x >= n)return true;
else return false;
}

int main()
{

while(scanf("%d %d %d", &n, &a, &b) != EOF)
{
if(n == a + b)
{
printf("1\n");
continue;
}
if(a > b)
{
int tmp = a;
a = b;
b = tmp;
}
if(b / a + 1 >= n)
{
printf("%d\n", a);
continue;
}
int l = 0, r = 100;
while(r - l > 1)
{
//printf("flag");
int mid = (l + r) / 2;
if(judge(mid))l = mid;
else r = mid;
}
printf("%d\n", l);
}
return 0;
}

C. Three Garlands

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.

When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if i-th garland is
switched on during x-th second, then it is lit only during seconds x, x + ki, x + 2ki, x + 3ki and
so on.

Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers x1, x2 and x3 (not
necessarily distinct) so that he will switch on the first garland during x1-th
second, the second one — during x2-th
second, and the third one — during x3-th
second, respectively, and during each second starting from max(x1, x2, x3) at
least one garland will be lit.

Help Mishka by telling him if it is possible to do this!

Input

The first line contains three integers k1, k2 and k3 (1 ≤ ki ≤ 1500)
— time intervals of the garlands.

Output

If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.

Otherwise, print NO.

Examples

input

2 2 3

output

YES

input

4 2 3

output

NO

Note

In the first example Mishka can choose x1 = 1, x2 = 2, x3 = 1.
The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ...,
which already cover all the seconds after the 2-nd one. It doesn't even matter what x3 is
chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.

In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit.

题解：在纸上画了一下，发现当最小的那个k大于3时无解，所以只要枚举最小的k分别等于1，2，3时的各种情况就好了
代码

#include<cstdio>
#include<algorithm>

using namespace std;

int a[3];

int main()
{
while(scanf("%d %d %d", &a[0], &a[1], &a[2]) != EOF)
{
sort(a, a + 3);
if(a[0] == 1)
{
printf("YES\n");
continue;
}
if(a[0] == 2)
{
if(a[1] == 2)
{
printf("YES\n");
continue;
}
if(a[1] == 4 && a[2] == 4)
{
printf("YES\n");
continue;
}
}
if(a[0] == 3)
{
if(a[1] == 3 && a[2] == 3)
{
printf("YES\n");
continue;
}
}
printf("NO\n");
}
return 0;
}