huangjing
思路：
就是（x,y）在两个参考系中的表示演完全一样。那么只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
Output
If they can meet with each other, please output "YES". Otherwise, please output "NO".
Sample Input
<p class="sycode"> 10 10 5 510 10 6 6 </p>
Sample Output
<p class="sycode"> YESNO </p>
Source
BestCoder Round #11 (Div. 2)
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 0x3f3f3f3fusing 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"); } }</queue></string></cmath></vector></map></algorithm></cstring></cstdio></iostream>题意：
就是给出n个数字，然后要你找到一个满足如下条件的数。
（1）这个数是奇数。
（2）这个数是是最大的数。
（3）还有一个被cha的点是所有的数字都要用到。我就是0 0 1 被cha了。。我还故意特判这种情况，都是题目没有读懂啊。。
思路：
贪心的做法，首先看所有的位是否存在基数，如果基数都没有，那么肯定是不存在这种数的，其次如果有，那么就将最小的基数找出来做各位，然后将所有的位进行排序，然后从低位向高位赋值，那么就得到这个树了，最后判断一下，如果首位为0，那么这个数就是不存在的，因为要求输出所有的位。。
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:
Input
There are multiple test cases. Please process till EOF.
Each case starts with a line containing an integer N ( 1 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
<p class="sycode"> 30 1 335 4 232 4 6 </p>
Sample Output
<p class="sycode"> 301425-1 </p>
Source
BestCoder Round #11 (Div. 2)
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 0x3f3f3f3fusing 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 <br> <br> <p class="sycode"> hdu 5056 Boring count </p> <p class="sycode"> 题意： </p> <p class="sycode"> 给出一个字符串，然后求出它所有的子串中每个字母的数目不超过k个的所有的子串的数目。。 </p> <p class="sycode"> 思路： </p> <p class="sycode"> 枚举每一个字符，然后以每个字符i为子串末尾，然后得到的满足条件的子串的最长长度。。就算字母相同，只要位置不相同就算不同的。。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，同时将当前的每个cnt[i]减减，直到移动到与i相同的字符你，那么从st到i这段字符就满足条件了。。。觉得这个思路真是神奇。。。。 </p> <p class="sycode"> 题目; </p> <p class="sycode"> Boring count <strong>Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)<br> Total Submission(s): 451 Accepted Submission(s): 169<br> </strong> <br> <br> </p>
<p class="sycode"> Problem Description </p> <p class="sycode"> 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. </p> <p class="sycode"> </p> <br> <p class="sycode"> Input </p> <p class="sycode"> In the first line there is an integer T , indicates the number of test cases. <br> For each case, the first line contains a string which only consist of lowercase letters. The second line contains an integer K. <br> <br> [Technical Specification] <br> 1 1 1 </p>
<p class="sycode"> </p> <br> <p class="sycode"> Output </p> <p class="sycode"> For each case, output a line contains the answer. </p> <p class="sycode"> </p> <br> <p class="sycode"> Sample Input </p> <p class="sycode"> </p>
<pre style="代码" class="precsshei"> <p class="sycode"> 3abc1abcabc1abcabc2 </p>
Sample Output
<p class="sycode"> 61521 </p>
Source
BestCoder Round #11 (Div. 2)
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 0x3f3f3f3fusing 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;}</queue></string></cmath></vector></map></algorithm></cstring></cstdio></iostream>
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