Educational Codeforces Round 23总结

A. Treasure Hunt

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Captain Bill the Hummingbird and his crew recieved an interesting challenge offer. Some stranger gave them a map, potion of teleportation and said that only this potion might help them to reach the treasure.

Bottle with potion has two values x and y written on it. These values define four moves which can be performed using the potion:

• 
• 
• 
• 

Map shows that the position of Captain Bill the Hummingbird is (x1, y1) and the position of the treasure is (x2, y2).

You task is to tell Captain Bill the Hummingbird whether he should accept this challenge or decline. If it is possible for Captain to reach the treasure using the potion then output "YES", otherwise "NO" (without quotes).

The potion can be used infinite amount of times.

Input

The first line contains four integer numbers x1, y1, x2, y2 ( - 105 ≤ x1, y1, x2, y2 ≤ 105) — positions of Captain Bill the Hummingbird and treasure respectively.

The second line contains two integer numbers x, y (1 ≤ x, y ≤ 105) — values on the potion bottle.

Output

Print "YES" if it is possible for Captain to reach the treasure using the potion, otherwise print "NO" (without quotes).

Examples

input

0 0 0 62 3

output

YES

input

1 1 3 61 5

output

NO

Note

In the first example there exists such sequence of moves:

1.  — the first type of move
2.  — the third type of move

题意：有4种变换方式，问最少需要多少步可以变到目标数字

解：判断是否可以解，可解的话求整除数

import java.util.Scanner;/** *  * 作者：张宇翔 创建日期：2017年6月16日 上午9:00:48 描述：写字楼里写字间，写字间里程序员； 程序人员写程序，又拿程序换酒钱。 * 酒醒只在网上坐，酒醉还来网下眠； 酒醉酒醒日复日，网上网下年复年。 但愿老死电脑间，不愿鞠躬老板前； 奔驰宝马贵者趣，公交自行程序员。 * 别人笑我忒疯癫，我笑自己命太贱； 不见满街漂亮妹，哪个归得程序员？ */public class Main {private static int x1,y1,x2,y2,x,y;public static void main(String[] args) {InitData();GetAns();}private static void InitData(){Scanner cin=new Scanner(System.in);x1=cin.nextInt();y1=cin.nextInt();x2=cin.nextInt();y2=cin.nextInt();x=cin.nextInt();y=cin.nextInt();};private static void GetAns(){int X=Math.abs(x1-x2);int Y=Math.abs(y1-y2);boolean ok=true;if(X%x!=0){ok=false;}if(Y%y!=0){ok=false;}if(!ok){System.out.println("NO");return;}else{int ans1=X/x;int ans2=Y/y;int ans=Math.abs(ans2-ans1);if(ans%2==0){System.out.println("YES");}else{System.out.println("NO");}}};}

B. Makes And The Product

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

After returning from the army Makes received a gift — an array a consisting of n positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i,  j,  k) (i < j < k), such that ai·aj·ak is minimum possible, are there in the array? Help him with it!

Input

The first line of input contains a positive integer number n (3 ≤ n ≤ 105) — the number of elements in array a. The second line containsn positive integer numbers ai (1 ≤ ai ≤ 109) — the elements of a given array.

Output

Print one number — the quantity of triples (i,  j,  k) such that i,  j and k are pairwise distinct and ai·aj·ak is minimum possible.

Examples

input

41 1 1 1

output

4

input

51 3 2 3 4

output

2

input

61 3 3 1 3 2

output

1

Note

In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.

In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.

In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.

题意：求数列中组成三个数的积最小的组合数个数

解：当最小数的个数大于3，有一个数学公式:(n)*(n+1)*(n+2)/6,但是我没推出来，就用前n项和。后面的情况见代码

import java.util.Arrays;import java.util.Scanner;/** *  * 作者：张宇翔 创建日期：2017年6月16日 上午9:00:48 描述：写字楼里写字间，写字间里程序员； 程序人员写程序，又拿程序换酒钱。 * 酒醒只在网上坐，酒醉还来网下眠； 酒醉酒醒日复日，网上网下年复年。 但愿老死电脑间，不愿鞠躬老板前； 奔驰宝马贵者趣，公交自行程序员。 * 别人笑我忒疯癫，我笑自己命太贱； 不见满街漂亮妹，哪个归得程序员？ */public class Main {private final static int Max = (int) (1e5 + 10);private static int n;private static int[] A;private static int ans1, ans2, ans3;private static int MIN1, MIN2, MIN3;public static void main(String[] args) {InitData();GetAns();}private static void InitData() {Scanner cin = new Scanner(System.in);n = cin.nextInt();A = new int[Max];ans1 = 0;ans2 = 0;ans3 = 0;MIN1 = -1;MIN2 = -1;MIN3 = -1;for (int i = 0; i < n; i++) {A[i] = cin.nextInt();}Arrays.sort(A, 0, n);};private static void GetAns() {MIN1 = A[0];ans1++;int i = 1;for (; i < n; i++) {if (A[i] == MIN1) {ans1++;} else {break;}}if (i < n) {MIN2 = A[i];i++;ans2++;for (; i < n; i++) {if (A[i] == MIN2) {ans2++;} else {break;}}if (i < n) {MIN3 = A[i];ans3++;i++;for (; i < n; i++) {if (A[i] == MIN3) {ans3++;} else {break;}}}}long sum=0;if(ans1>=3){ans1-=2;for(long j=1;j<=ans1;j++){sum+=((j*(j+1))/2);}}else{if(ans1==2){sum=ans2;}else{if(ans2==2){sum=1;}else{sum=ans3;}}}System.out.println(sum);};}

C. Really Big Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number x is really big if the difference between x and the sum of its digits (in decimal representation) is not less than s. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than n.

Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.

Input

The first (and the only) line contains two integers n and s (1 ≤ n, s ≤ 1018).

Output

Print one integer — the quantity of really big numbers that are not greater than n.

Examples

input

12 1

output

3

input

25 20

output

0

input

10 9

output

1

Note

In the first example numbers 10, 11 and 12 are really big.

In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30:30 - 3 ≥ 20).

In the third example 10 is the only really big number (10 - 1 ≥ 9).

题意：找一个数n，ans为n的各个位之和，n-ans>=s才能称为大数；

思路：只需要保证大数个位是9，大数后面的数字就全部满足条件了，从s循环至n，结束标志便是个位是9的大数；

import java.util.Scanner;/** *  * 作者：张宇翔 创建日期：2017年6月16日 上午9:00:48 描述：写字楼里写字间，写字间里程序员； 程序人员写程序，又拿程序换酒钱。 * 酒醒只在网上坐，酒醉还来网下眠； 酒醉酒醒日复日，网上网下年复年。 但愿老死电脑间，不愿鞠躬老板前； 奔驰宝马贵者趣，公交自行程序员。 * 别人笑我忒疯癫，我笑自己命太贱； 不见满街漂亮妹，哪个归得程序员？ */public class Main {private final static int Max = (int) (1e5 + 10);private static long n,s;public static void main(String[] args) {InitData();GetAns();}private static void InitData() {Scanner cin = new Scanner(System.in);n=cin.nextLong();s=cin.nextLong();};private static void GetAns() {long i;long ans=0;for(i=s;i<=n;i++){long k=i-sum(i);if(k>=s){if(i%10==9){break;}ans++;}}if(n>=s){System.out.println(ans-i+n+1);}else{System.out.println(0);}};private static long sum(long ans){long sum=0;while(ans>0){sum+=(ans%10);ans/=10;}return sum;}}