hihocoder-Weekly224-Split Array

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hihocoder-Weekly224-Split Array

 

題目1 : Split Array時間限制:10000ms單點時限:1000ms記憶體限制: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}.  

輸入

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)

輸出

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

範例輸入

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

範例輸出

YES  NO

 

 

題解：

　　很簡單的題目

　　因為該數組是已經排好序的，而且其範圍是1-10^6 , 可以用一個數組記錄每一個數的cnt，然後順序便利一遍，就可以判斷。

　　時間複雜度： 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\n");    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\n");    }else{    printf("NO\n");    }    } } 

　　

 

hihocoder-Weekly224-Split Array