ZOJ1569

n m
a1 a2 a3 a4 a5 a6 a7 a8 ...
S1 = a1
S2 = a1 + a2
S3 = a1 + a2 + a3
S4 = a1 + a2 + a3 + a4
S5 = a1 + a2 + a3 + a4 + a5

任何一段连续数和可用Sj - Si (i < j)求得
若Sj % m == Si % m,可以知道a[i] + a[i+1] + ...+ a[j]%m == 0,
即此段和可以被m除尽 (对零做特殊处理)

 

#include <iostream>
using namespace std;

const int MAX_SIZE = 5005;
//这里有问题，当改为5000时，提交一直是Segmentation Fault 
//而照题目所言 ，m (m <= 5000).汗！ 
int Count[MAX_SIZE];
int n,m;

void Init()
{
    for(int i = 0; i < m; i++)
        Count[i] = 0;//S[x] % m 为m，其计数为这样的部分和个数
}

int main()
{
    int cnt,sum,tmp;;
    while(cin>>n>>m)
    {
        cnt = 0;
        sum = 0;
        Init();
        while(n--)
        {
            cin>>tmp;
            sum =(sum + tmp)%m;
            if(!sum)cnt++;//对零特殊处理
            Count[sum]++;
        }
        for(int i = 0; i < m; i++)
            cnt += Count[i]*(Count[i] - 1) / 2;
        cout<<cnt<<endl;
    }
    return 0;
}

 

Segmentation Fault :

• 1.buffer overflow --- usually caused by a pointer reference out of range.
  
• 2.stack overflow --- please keep in mind that the default stack size is 8192K.