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BestCoderRound#11(Div.2)前三題題解_html/css

來源：懂視網責編：小采時間：2020-11-27 15:55:59

BestCoderRound#11(Div.2)前三題題解_html/css

BestCoderRound#11(Div.2)前三題題解_html/css_WEB-ITnose:題目鏈接： huangjing hdu5054 Alice and Bob 思路：就是（x,y）在兩個參考系中的表示演完全一樣。那么只可能在這個矩形的中點。題目: Alice and Bob Time Limit: 2000/1000 MS (Java/Others) Memory Li

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導讀BestCoderRound#11(Div.2)前三題題解_html/css_WEB-ITnose:題目鏈接： huangjing hdu5054 Alice and Bob 思路：就是（x,y）在兩個參考系中的表示演完全一樣。那么只可能在這個矩形的中點。題目: Alice and Bob Time Limit: 2000/1000 MS (Java/Others) Memory Li

題目鏈接：

huangjing

hdu5054 Alice and Bob

思路：

就是（x,y）在兩個參考系中的表示演完全一樣。那么只可能在這個矩形的中點。

題目:

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 5 510 10 6 6

Sample Output

YESNO

Source

BestCoder Round #11 (Div. 2)

Recommend

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

代碼：

#include#include#include#include#include#include#include#include#include#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"); } }

hdu 5055 Bob and math problem

題意：

就是給出n個數字，然后要你找到一個滿足如下條件的數。

（1）這個數是奇數。

（2）這個數是是最大的數。

（3）還有一個被cha的點是所有的數字都要用到。我就是0 0 1 被cha了。我還故意特判這種情況，都是題目沒有讀懂啊。

思路：

貪心的做法，首先看所有的位是否存在基數，如果基數都沒有，那么肯定是不存在這種數的，其次如果有，那么就將最小的基數找出來做各位，然后將所有的位進行排序，然后從低位向高位賦值，那么就得到這個樹了，最后判斷一下，如果首位為0，那么這個數就是不存在的，因為要求輸出所有的位。

題目：

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

30 1 335 4 232 4 6

Sample Output

301425-1

Source

BestCoder Round #11 (Div. 2)

Recommend

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

代碼：

#include#include#include#include#include#include#include#include#include#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<=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

題意：

給出一個字符串，然后求出它所有的子串中每個字母的數目不超過k個的所有的子串的數目。

思路：

枚舉每一個字符，然后以每個字符i為子串末尾，然后得到的滿足條件的子串的最長長度。就算字母相同，只要位置不相同就算不同的。2333333333，那么思路就是維護一個起點st，每當第i個字符的數目大于k后，那么就將st后移，同時將當前的每個cnt[i]減減，直到移動到與i相同的字符你，那么從st到i這段字符就滿足條件了。。覺得這個思路真是神奇。。

題目;

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

3abc1abcabc1abcabc2

Sample Output

61521

Source

BestCoder Round #11 (Div. 2)

Recommend

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

代碼：

#include#include#include#include#include#include#include#include#include#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;}

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BestCoderRound#11(Div.2)前三題題解_html/css

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BestCoderRound#11(Div.2)前三題題解_html/css

BestCoderRound#11(Div.2)前三題題解_html/css

題目鏈接：

hdu5054 Alice and Bob

BestCoderRound#11(Div.2)前三題題解_html/css

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