huangjing

# Alice and Bob

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 216    Accepted Submission(s): 166

Problem Description

Bob and Alice got separated in the Square, they agreed that if they get separated, they‘ll meet back at the coordinate point (x, y). Unfortunately they forgot to define the origin of coordinates and the coordinate axis direction. Now, Bob in the lower left
corner of the Square, Alice in the upper right corner of the the Square. Bob regards the lower left corner as the origin of coordinates, rightward for positive direction of axis X, upward for positive direction of axis Y. Alice regards the upper right corner
as the origin of coordinates, leftward for positive direction of axis X, downward for positive direction of axis Y. Assuming that Square is a rectangular, length and width size is N * M. As shown in the figure:

Bob and Alice with their own definition of the coordinate system respectively, went to the coordinate point (x, y). Can they meet with each other ?

Note: Bob and Alice before reaching its destination, can not see each other because of some factors (such as buildings, time poor).

Input

There are multiple test cases. Please process till EOF. Each test case only contains four integers : N, M and x, y. The Square size is N * M, and meet in coordinate point (x, y). ( 0 < x < N <= 1000 , 0 < y < M <= 1000 ).

Output

If they can meet with each other, please output "YES". Otherwise, please output "NO".

Sample Input

10 10 5 5
10 10 6 6


Sample Output

YES
NO


Source

BestCoder Round #11 (Div. 2)

Recommend

heyang   |   We have carefully selected several similar problems for you:  5057 5052 5051 5050 5049

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int  main()
{
int x,y,n,m;
while(scanf("%d%d%d%d",&n,&m,&x,&y)!=EOF)
{
if(2*x==n&&2*y==m)
printf("YES\n");
else
printf("NO\n");
}
}

hdu 5055 Bob and math problem

（1）这个数是奇数。

（2）这个数是是最大的数。

（3）另一个被cha的点是全部的数字都要用到。我就是0 0 1 被cha了。。我还有益特判这样的情况，都是题目没有读懂啊。。

# Bob and math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 643    Accepted Submission(s): 245

Problem Description

Recently, Bob has been thinking about a math problem.

There are N Digits, each digit is between 0 and 9. You need to use this N Digits to constitute an Integer.

This Integer needs to satisfy the following conditions:

• 1. must be an odd Integer.
• 2. there is no leading zero.
• 3. find the biggest one which is satisfied 1, 2.

Example:

There are three Digits: 0, 1, 3. It can constitute six number of Integers. Only "301", "103" is legal, while "130", "310", "013", "031" is illegal. The biggest one of odd Integer is "301".

Input

There are multiple test cases. Please process till EOF.

Each case starts with a line containing an integer N ( 1 <= N <= 100 ).

The second line contains N Digits _1, a_2, a_3, \cdots, a_n. ( 0 \leqwhich indicate the digit $a a_i \leq 9)$.

Output

The output of each test case of a line. If you can constitute an Integer which is satisfied above conditions, please output the biggest one. Otherwise, output "-1" instead.

Sample Input

3
0 1 3
3
5 4 2
3
2 4 6


Sample Output

301
425
-1


Source

BestCoder Round #11 (Div. 2)

Recommend

heyang   |   We have carefully selected several similar problems for you:  5057 5052 5051 5050 5049

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=100+10;
int a[maxn],odd[maxn];
char str[maxn];
int n;

int  main()
{
int ans,pd;
while(scanf("%d",&n)!=EOF)
{
memset(str,0,sizeof(str));
int cnt=0,first=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]%2)
odd[++cnt]=a[i];
}
sort(odd+1,odd+1+cnt);
sort(a+1,a+1+n);
int ly=n-1;
if(cnt==0)
puts("-1");
else
{
str[ly]=odd[1]+'0';
ly--;
for(int i=1;i<=n;i++)
{
if(a[i]==odd[1]&&!first)
{
first=1;
continue;
}
else
{
str[ly]=a[i]+'0';
ly--;
}
}
if(str[0]=='0')
puts("-1");
else
{
for(int i=0;i<=n-1;i++)
printf("%c",str[i]);
printf("\n");
}
}
}
return 0;
}


hdu 5056 Boring count

# Boring count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 451    Accepted Submission(s): 169

Problem Description

You are given a string S consisting of lowercase letters, and your task is counting the number of substring that the number of each lowercase letter in the substring is no more than K.

Input

In the first line there is an integer T , indicates the number of test cases.

For each case, the first line contains a string which only consist of lowercase letters. The second line contains an integer K.

[Technical Specification]

1<=T<= 100

1 <= the length of S <= 100000

1 <= K <= 100000

Output

For each case, output a line contains the answer.

Sample Input

3
abc
1
abcabc
1
abcabc
2


Sample Output

6
15
21


Source

BestCoder Round #11 (Div. 2)

Recommend

heyang   |   We have carefully selected several similar problems for you:  5057 5052 5051 5050 5049

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const  int maxn=100000+10;
char str[maxn];
int cnt[28];

int main()
{
ll ans;
int t,st,k,ly;
scanf("%d",&t);
while(t--)
{
memset(cnt,0,sizeof(cnt));
st=ans=0;
scanf("%s%d",str,&k);
for(int i=0;str[i]!='\0';i++)
{
ly=str[i]-'a';
cnt[ly]++;
if(cnt[ly]>k)
{
while(str[st]!=str[i])
{
cnt[str[st]-'a']--;
st++;
}
cnt[ly]--;
st++;
}
ans+=i-st+1;
}
printf("%I64d\n",ans);
}
return 0;
}


## BestCoder Round #11 (Div. 2) 题解

HDOJ5054 Alice and Bob Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 302    Accepted Submission(s): 229 Problem Description Bob and Alice got separated in the Square, they agreed that if they

## HDU 5056 Boring count（BestCoder Round #11 (Div. 2)）

Problem Description: You are given a string S consisting of lowercase letters, and your task is counting the number of substring that the number of each lowercase letter in the substring is no more than K. Input: In the first line there is an integer

## hdu5418 BestCoder Round #52 (div.2) Victor and World （ floyd+状压dp）

Problem Description After trying hard for many years, Victor has finally received a pilot license. To have a celebration, he intends to buy himself an airplane and fly around the world. There are n countries on the earth, which are numbered from 1 to

## BestCoder Round #69 (div.2) Baby Ming and Weight lifting（hdu 5610）

Baby Ming and Weight lifting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 681    Accepted Submission(s): 280 Problem Description Baby Ming is fond of weight lifting. He has a barbell pole(the