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
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).
10 10 5 5
10 10 6 6
YES
NO
/**
* Created by ckboss on 14-10-3.
*/
import java.util.*; public class Main {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
while(in.hasNext())
{
int N=in.nextInt();
int M=in.nextInt();
int x=in.nextInt();
int y=in.nextInt();
if(x==N-x&&y==M-y)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
HDOJ5055 Bob and math problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 812 Accepted Submission(s): 313
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".
Each case starts with a line containing an integer N ( 1 <= N <= 100 ).
The second line contains N Digits which indicate the digit $a_1, a_2, a_3, \cdots, a_n. ( 0 \leq a_i \leq 9)$.
3
0 1 3
3
5 4 2
3
2 4 6
301
425
-1
/**
* Created by ckboss on 14-10-3.
*/
import java.util.*; public class Main { static int[] num = new int[10];
static char[] ans = new char[110];
static int nt;
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int n;
while(in.hasNext()){
n=in.nextInt();
int flag=0;
int MinJ=111;
Arrays.fill(num,0); nt=0;
for(int i=0;i<n;i++){
int x=in.nextInt();
num[x]++;
if(x%2==1){
if(x<MinJ) MinJ=x;
flag++;
}
}
if(flag==0)
{
System.out.println("-1");
continue;
}
num[MinJ]--;
for(int i=9;i>=0;i--){
for(int j=num[i];j>0;j--){
ans[nt++]=(char)('0'+i);
}
}
ans[nt++]=(char)(MinJ+'0');
flag=0;
for(int i=0;i<nt;i++){
if(ans[i]=='0' && flag==0){
flag=0; break;
}
flag=1;
System.out.print(ans[i]);
}
if(flag==1)
System.out.println("");
else
System.out.println("-1");
}
}
}
HDOJ5056 Boring count
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 615 Accepted Submission(s): 242
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
3
abc
1
abcabc
1
abcabc
2
6
15
21
O(n)贪心...
/**
* Created by ckboss on 14-10-3.
*/
import java.util.*; public class Main { static long[] num = new long[30]; public static void main(String[] args){
Scanner in = new Scanner(System.in);
int T_T=in.nextInt();
while(T_T-->0){
String st=in.next();
int k=in.nextInt();
Arrays.fill(num,0);
int be=0,ed=0;
long ans=0;
for(int i=0,sz=st.length();i<sz;i++){
int id=(int)(st.charAt(i)-'a');
num[id]++;
if(num[id]>k){
for(;be<=ed;be++){
ans+=ed-be+1;
num[st.charAt(be)-'a']--;
if(st.charAt(be)-'a'==id) {
be++;
break;
}
}
}
ed=i;
}
for(;be<=ed;be++){
ans+=ed-be+1;
num[st.charAt(be)-'a']--;
}
System.out.println(ans);
}
}
}
HDOJ5057 Argestes and Sequence
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 603 Accepted Submission(s): 152
can be one of the following:
S X Y: you should set the value of a[x] to y(in other words perform an assignment a[x]=y).
Q L R D P: among [L, R], L and R are the index of the sequence, how many numbers that the Dth digit of the numbers is P.
Note: The 1st digit of a number is the least significant digit.
For each case, the first line contains two numbers N and M.The second line contains N integers, separated by space: a[1],a[2],...,a[n]—initial value of array elements.
Each of the next M lines begins with a character type.
If type==S,there will be two integers more in the line: X,Y.
If type==Q,there will be four integers more in the line: L R D P.
[Technical Specification]
1<=T<= 50
1<=N, M<=100000
0<=a[i]<=$2^{31}$ - 1
1<=X<=N
0<=Y<=$2^{31}$ - 1
1<=L<=R<=N
1<=D<=10
0<=P<=9
1
5 7
10 11 12 13 14
Q 1 5 2 1
Q 1 5 1 0
Q 1 5 1 1
Q 1 5 3 0
Q 1 5 3 1
S 1 100
Q 1 5 3 1
5
1
1
5
0
1
分块大法好....
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath> using namespace std; const int maxn=100100; struct BLOCK
{
int cnt[10][10];
}block[400];
int block_size,block_num;
int n,m;
int a[maxn]; const int ten[12]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000}; void CHANGE(int p,int v)
{
int id=p/block_size;
int x=a[p];
for(int i=0;i<10;i++)
{
block[id].cnt[i][x%10]--;
x/=10;
}
a[p]=v;
x=v;
for(int i=0;i<10;i++)
{
block[id].cnt[i][x%10]++;
x/=10;
}
} int QUERY(int l,int r,int p,int d)
{
int L=l/block_size,R=r/block_size;
int ans=0;
if(R-L<=1)
{
for(int i=l;i<=r;i++)
{
ans+=((a[i]/ten[p-1])%10==d)?1:0;
}
return ans;
}
for(int i=l;i<(L+1)*block_size;i++)
{
ans+=((a[i]/ten[p-1])%10==d)?1:0;
}
for(int i=L+1;i<=R-1;i++)
{
ans+=block[i].cnt[p-1][d];
}
for(int i=R*block_size;i<=r;i++)
{
ans+=((a[i]/ten[p-1])%10==d)?1:0;
}
return ans;
} int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&m);
memset(block,0,sizeof(block));
block_size=sqrt(n*1.0)+1;
block_num=n/block_size+1;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
int x=a[i];
int id=i/block_size;
for(int j=0;j<10;j++)
{
block[id].cnt[j][x%10]++;
x/=10;
}
}
while(m--)
{
char op[10];
int a,b,c,d;
scanf("%s",op);
if(op[0]=='Q')
{
scanf("%d%d%d%d",&a,&b,&c,&d);
printf("%d\n",QUERY(a,b,c,d));
}
else if(op[0]=='S')
{
scanf("%d%d",&a,&b);
CHANGE(a,b);
}
}
}
return 0;
}
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