题目链接：
hdu5054 Alice and Bob
思路：
就是（x,y）在两个參考系中的表示演全然一样。那么仅仅可能在这个矩形的中点。。
题目:
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
Recommend
heyang | We have carefully selected several similar problems for you: 5057 5052 5051 5050 5049
代码：
#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
题意：
就是给出n个数字，然后要你找到一个满足例如以下条件的数。
（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。。
思路：
贪心的做法，首先看全部的位是否存在基数，假设基数都没有，那么肯定是不存在这样的数的，其次假设有，那么就将最小的基数找出来做各位，然后将全部的位进行排序，然后从低位向高位赋值，那么就得到这个树了，最后推断一下，假设首位为0，那么这个数就是不存在的，由于要求输出全部的位。。
题目：
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
Recommend
heyang | We have carefully selected several similar problems for you: 5057 5052 5051 5050 5049
代码：
#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
题意：
给出一个字符串，然后求出它全部的子串中每一个字母的数目不超过k个的全部的子串的数目。。
思路：
枚举每一个字符，然后以每一个字符i为子串末尾，然后得到的满足条件的子串的最长长度。。就算字母同样，仅仅要位置不同样就算不同的。。2333333333，那么思路就是维护一个起点st，每当第i个字符的数目大于k后，那么就将st后移，同一时候将当前的每一个cnt[i]减减，直到移动到与i同样的字符你，那么从st到i这段字符就满足条件了。。。认为这个思路真是奇妙。。。。
题目;
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
Recommend
heyang | We have carefully selected several similar problems for you: 5057 5052 5051 5050 5049
代码：
#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;
}