BestCoder Round #36 ($)（一、二题解题报告）

Strange Class

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 89    Accepted Submission(s): 71

Problem Description

In Vivid’s school, there is a strange class(SC). In SC, the students’ names are very strange. They are in the same format: anbncn(a,b,c must not be the same with each other). For example studens whose names are“abc”,”ddppqq” are in SC, however studens whose names are “aaa”,“ab”,”ddppqqq” are not in SC.
Vivid makes friends with so many students, he wants to know who are in SC.

 

Input

There are multiple test cases (about 10), each case will give a string S which is the name of Vivid’s friend in a single line.
Please process to the end of file.

[Technical Specification]

1≤|S|≤10.

|S| indicates the length of S.

S only contains lowercase letter.

 

Output

For each case, output YES if Vivid’s friend is the student of SC, otherwise output NO.

 

Sample Input

abcbc

 

Sample Output

YESNO

 

Source

BestCoder Round #36 ($)

 

解题思想：

这是一个简单题，先判断长度是不是3的倍数。然后再判断前中后三段里面的字符是不是一样。最后判断一下这三段之间的字符是不一样的。

AC  code：

#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<string.h>#include<string>#define MAXN 1000010#define LL long longusing namespace std;int main(){string s;int i,j,k,fg,t1,t2;while(cin>>s){if(s.size()%3!=0){printf("NO\n");}else{if(s.size()==3){if(s[0]!=s[1]&&s[1]!=s[2]&&s[0]!=s[3]){printf("YES\n");}else{printf("NO\n");}}else{fg=1;for(i=1;i<s.size()/3;i++){if(s[i]!=s[i-1]){fg=0;break;}}for(j=i+1;fg&&j<i+s.size()/3;j++){if(s[j]!=s[j-1]){fg=0;break;}}for(k=j+1;fg&&k<s.size();k++){if(s[k]!=s[k-1]){fg=0;break;}}if(fg){if(s[i-1]!=s[j-1]&&s[j-1]!=s[k-1]&&s[k-1]!=s[i-1]){fg=1;}else{fg=0;}}if(fg){printf("YES\n");}else{printf("NO\n");}}}}return 0; } 

Gunner

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 484    Accepted Submission(s): 223

Problem Description

Long long ago, there is a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds and n trees. The i−th bird stands on the top of the i−th tree. The trees stand in straight line from left to the right. Every tree has its height. Jack stands on the left side of the left most tree. When Jack shots a bullet in height H to the right, the bird which stands in the tree with height H will falls.
Jack will shot many times, he wants to know how many birds fall during each shot.

a bullet can hit many birds, as long as they stand on the top of the tree with height of H.

 

Input

There are multiple test cases (about 5), every case gives n,m in the first line, n indicates there are n trees and n birds, m means Jack will shot m times.

In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which describes the height of the trees.

In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which describes the height of the Jack’s shots.

Please process to the end of file.

[Technical Specification]

1≤n,m≤1000000(106)

1≤h[i],q[i]≤1000000000(109)

All inputs are integers.

 

Output

For each q[i], output an integer in a single line indicates the number of birds Jack shot down.

 

Sample Input

4 31 2 3 41 1 4

 

Sample Output

101
Hint
Huge input, fast IO is recommended.
 

 

Source

BestCoder Round #36 ($)

 

解题思想：

先对数据进行排序，然后对每一个查询进行二分查找。注意查找到之后要删除。还有一种做法是用hash，两个算法的效率差不多。

注意：这题卡时间卡得要死，输出0时不能用printf，否则TLE！！！只能用puts输出。。。

AC  code：

/*printf("%s\n",s);puts(s);两个函数在跑起来哪个快在数据量 很大的时候应该选哪个（都是输出字符串的）是puts()函数，因为它不需要去解析格式控制符，省去了这个额外的工作。*/#include<iostream>#include<stdio.h>#include<map>#include<vector>#include<set>#include<cstdlib>#include<string.h>#include<algorithm>#include<cmath>#define MAXN 1000010using namespace std;struct node{int val;int num;}a[MAXN];int cnt;int h[MAXN],q;bool vis[MAXN];bool cmp(node a,node b){return a.val<b.val;}int read(){int fg=1,ret=0;char ch;ch=getchar();while(!(ch>='0'&&ch<='9')){if(ch=='-')fg=-1;ch=getchar();}while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}return ret*fg;}int er(int x){int l=1,r=cnt,mid;while(l<=r){mid=(l+r)/2;if(a[mid].val==x){return mid;}else if(a[mid].val>x){r=mid-1;}else{l=mid+1;}}return 0;}int main(){int n,m,i;while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;i++){h[i]=read();}sort(h+1,h+n+1);cnt=1;a[cnt].num=1;a[cnt].val=h[1];for(i=2;i<=n;i++){if(h[i]==h[i-1]){a[cnt].num++;}else{a[++cnt].num=1;a[cnt].val=h[i];}}memset(vis,0,sizeof(vis));while(m--){q=read();int id=er(q);if(vis[id]||id==0){puts("0");//printf("0\n");//不能用printf，因为puts更快，不会TLE！！！ }else{printf("%d\n",a[id].num);vis[id]=1;}}}return 0;}