HDU 4648 Magic Pen 6

Magic Pen 6

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 399    Accepted Submission(s): 166

Problem Description

In HIT, many people have a magic pen. Lilu0355 has a magic pen, darkgt has a magic pen, discover has a magic pen. Recently, Timer also got a magic pen from seniors.

At the end of this term, teacher gives Timer a job to deliver the list of N students who fail the course to dean's office. Most of these students are Timer's friends, and Timer doesn't want to see them fail the course. So, Timer decides to use his magic pen to scratch out consecutive names as much as possible. However, teacher has already calculated the sum of all students' scores module M. Then in order not to let the teacher find anything strange, Timer should keep the sum of the rest of students' scores module M the same.

Plans can never keep pace with changes, Timer is too busy to do this job. Therefore, he turns to you. He needs you to program to "save" these students as much as possible.

 

Input

There are multiple test cases.
The first line of each case contains two integer N and M, (0< N <= 100000, 0 < M < 10000),then followed by a line consists of N integers a ₁,a ₂,...a _n (-100000000 <= a ₁,a ₂,...a _n <= 100000000) denoting the score of each student.(Strange score? Yes, in great HIT, everything is possible)

 

Output

For each test case, output the largest number of students you can scratch out.

 

Sample Input

  
  

   
   2 3
1 6
3 3
2 3 6
2 5
1 3
  
  

 

Sample Output

  
  

   
   1
2
0

   
   

    
    

     
     Hint
    
    
The magic pen can be used only once to scratch out consecutive students.

   
   
   
    
  
  

 

Source

2013 Multi-University Training Contest 5

 

Recommend

zhuyuanchen520

 

题意：有N个数， 取出一个区间的数。  该区间的和要能整除M。  求最大的区间

思路： 水过了。O(n ^ 2)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

//

const int V = 100000 + 50;
const int MaxN = 80 + 5;
const int mod = 10000 + 7;
const __int64 INF = 0x7FFFFFFFFFFFFFFFLL;
const int inf = 0x7fffffff;
int n, m, sum[V], ans;
int main() {
    int i, j;
    while(~scanf("%d%d", &n, &m)) {
        ans = 0;
        for(i = 1; i <= n; ++i) {
            int temp;
            scanf("%d", &temp);
            sum[i] = (sum[i - 1] + temp) % m;
        }
        for(i = n; i >= 1 && i > ans; --i) {
            for(j = 1; j + i - 1 <= n; ++j)
                if((sum[j + i - 1] - sum[j - 1]) % m == 0) {
                    ans = max(ans, i);
                    break;
                }
        }
        printf("%d\n", ans);
    }
}

方法二：  贪心  O(NlogN) 

比上面的N^2慢， 不知道是数据水， 还是可以证明区间一定很接近N.

题目是要求  (sum[i] - sum[j] )% m == 0

那么只需要将每个 sum[i] % m 然后排序。 判断该模的最大区间。

这里的模一定是非负数， 坑了我很久   -2 % 8 = 6  而不是 -2.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

//

const int V = 100000 + 50;
const int MaxN = 80 + 5;
const int mod = 10000 + 7;
const __int64 INF = 0x7FFFFFFFFFFFFFFFLL;
const int inf = 0x7fffffff;
pair<int, int> pa[V];
int n, m, ans;
int main() {
    int i, j;
    while(~scanf("%d%d", &n, &m)) {
        ans = 0;
        pa[0].first = pa[0].second = 0;
        for(i = 1; i <= n; ++i) {
            int temp;
            scanf("%d", &temp);
            pa[i].first = (pa[i - 1].first + temp) % m;
            pa[i].first = (pa[i].first + m) % m;
            pa[i].second = i;
        }
        sort(pa, pa + n + 1);
        j = 0;
        for(i = 1; i <= n; ++i) {
            if(pa[i].first != pa[j].first)
                j = i;
            ans = max(ans, pa[i].second - pa[j].second);
        }
        printf("%d\n", ans);
    }
}

方法三：  O(N)

但是还是没有方法一快。无解。

每次只更新模的左右区间。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

//

const int V = 100000 + 50;
const int MaxN = 80 + 5;
const int mod = 10000 + 7;
const __int64 INF = 0x7FFFFFFFFFFFFFFFLL;
const int inf = 0x7fffffff;
int n, m, ans, sum[V];
pair<int, int> Mod[V]; // 模i的区间  [Mod[i].first,  Mod[i].second]
bool vis[V];
int main() {
    int i, j;
    while(~scanf("%d%d", &n, &m)) {
        ans = 0;
        memset(Mod, 0, sizeof(Mod));
        memset(vis, false, sizeof(vis));
        vis[0] = true;
        for(i = 1; i <= n; ++i) {
            int temp;
            scanf("%d", &temp);
            sum[i] = (sum[i - 1] + temp) % m;
            sum[i] = (sum[i] + m) % m;
            if(!vis[sum[i]]) {
                vis[sum[i]] = true;
                Mod[sum[i]].first = i;
            }
            else {
                Mod[sum[i]].second = max(Mod[sum[i]].second, i);
                ans = max(ans, Mod[sum[i]].second - Mod[sum[i]].first);
            }
        }
        printf("%d\n", ans);
    }
}