ZOJ 1569 Partial Sums

最新推荐文章于 2020-03-28 23:06:16 发布

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本文介绍了一种算法，用于解决给定一系列整数，找出所有可能的连续子序列，使得这些子序列的和能被另一个给定的整数整除的问题。通过巧妙地利用前缀和的概念及余数的特性，该算法能够有效地计算出满足条件的子序列数量。

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Partial Sums

Time Limit: 2 Seconds Memory Limit: 65536 KB

Given a series of n numbers a1, a2, ..., an, the partial sum of the numbers is defined as the sum of ai, ai+1, ..., aj.

You are supposed to calculate how many partial sums of a given series of numbers could be divided evenly by a given number m.

Input

There are multiple test, each contains 2 lines.

The first line is 2 positive integers n (n <= 10000 ) and m (m <= 5000).

The second line contains n non-negative integers a1, a2, ..., an. Numbers are separated by one or several spaces.

The input is ended by EOF.

Output

One test each line - the number of partial sums which could be divided by m.

Sample Input

5 4
1 2 3 4 5
6 7
9 8 7 6 5 4

Sample Output

2
3

题目大意：从n个数中选一些，他们的和可以被m整除，求有多少组这样的数。

sum[i...j]=sum[1....j]-sum[1....i-1]

如果sum[i...j]%m==0 >>> (sum[1...j]-sum[1...i-1])%m==0 >>> sum[1...j]%m==sum[1...i-1]%m

所以记录下从1...i的和存到sum数组中，若某个个sum[i]有K个，就有C(K,2)种结果

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m;
int a[10005];
int main()
{
    while(cin>>n>>m)
    {
        int ans=0,sum=0,x;
        memset(a,0,sizeof(a));
        for(int i=0; i<n; i++)
        {
            cin>>x;
            sum=(sum+x)%m;
            if(sum==0)
                ans++;
            a[sum]++;
        }
        for(int i=0; i<m; i++)
            if(a[i])
                ans+=a[i]*(a[i]-1)/2;
        cout<<ans<<endl;
    }
}