hihocoder-Weekly224-Split Array

hihocoder-Weekly224-Split Array

题目1 : Split Array

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

You are given an sorted integer array A and an integer K. Can you split A into several sub-arrays that each sub-array has exactly K continuous increasing integers.

For example you can split {1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6}  into {1, 2, 3}, {1, 2, 3}, {3, 4, 5}, {4, 5, 6}.  

输入

The first line contains an integer T denoting the number of test cases. (1 <= T <= 5)

Each test case takes 2 lines. The first line contains an integer N denoting the size of array A and an integer K. (1 <= N <= 50000, 1 <= K <= N)

The second line contains N integers denoting array A. (1 <= Ai <= 100000)

输出

For each test case output YES or NO in a separate line.

样例输入

2  
12 3 
1 1 2 2 3 3 3 4 4 5 5 6  
12 4  
1 1 2 2 3 3 3 4 4 5 5 6

样例输出

YES  
NO

题解：

　　很简单的题目

　　因为该数组是已经排好序的，而且其范围是1-10^6 , 可以用一个数组记录每一个数的cnt，然后顺序便利一遍，就可以判断。

　　时间复杂度： O(10^6)  

#include <cstdio>  
#include <cstring> 
#include <cstdlib> 
const int MAXN = 100000 + 10; 

#define max(a, b) (a)>(b)?(a):(b) 
#define min(a, b) (a)>(b)?(b):(a) 

int n, k, num[MAXN]; 

int main(){ 

    int test_case;
    scanf("%d", &test_case);  
    int x, min_x, max_x, tmp;   
    bool ans; 
    while(test_case--)
    {
    	memset(num, 0, sizeof(num)); 
    	scanf("%d %d", &n, &k);  
    	max_x = 0; min_x = MAXN; 
    	for(int i=0; i<n; ++i)
    	{
    		scanf("%d", &x); 
    		num[x]++;  
    		max_x = max(max_x, x);  
    		min_x = min(min_x, x); 
    	} 
    	if(n%k != 0){
    		printf("NO
");
    		continue; 
    	}  
    	ans = true; 
    	for(int i=min_x; i<=max_x; ++i)
    	{
    		if(num[i] > 0)
    		{
    			tmp = num[i]; 
    			for(int j=0; j<k; ++j)
    			{
    				if(i+j > max_x){
    					ans = false; 
    					break; 
    				}
    				num[i+j] -= tmp; 
    			}
    		}else if(num[i] < 0)
    		{
    			ans = false; 
    			break; 
    		}
    	}
    	if(ans){
    		printf("YES
");
    	}else{
    		printf("NO
");
    	}
    } 
}