AtCoder - 4240 Candy Distribution

最新推荐文章于 2024-10-25 20:34:43 发布

转载最新推荐文章于 2024-10-25 20:34:43 发布 · 495 阅读

数论专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种利用前缀和与取余数技巧解决特定数学问题的方法，核心在于通过计算数列的前缀和并对其取余，找到满足条件的数对，以确定数列中某段连续元素的和是否为给定数的倍数。

Problem Statement

There are N boxes arranged in a row from left to right. The i-th box from the left contains Ai candies.

You will take out the candies from some consecutive boxes and distribute them evenly to M children.

Such being the case, find the number of the pairs (l,r) that satisfy the following:

l and r are both integers and satisfy 1≤l≤r≤N.
Al+Al+1+…+Ar is a multiple of M.

Constraints

All values in input are integers.
1≤N≤105
2≤M≤109
1≤Ai≤109

Input

Input is given from Standard Input in the following format:

N M
A1 A2 … AN

Output

Print the number of the pairs (l,r) that satisfy the conditions.

Note that the number may not fit into a 32-bit integer type.

Sample Input 1

3 2
4 1 5

Sample Output 1

The sum Al+Al+1+…+Ar for each pair (l,r) is as follows:

Sum for (1,1): 4
Sum for (1,2): 5
Sum for (1,3): 10
Sum for (2,2): 1
Sum for (2,3): 6
Sum for (3,3): 5

Among these, three are multiples of 2.

Sample Input 2

13 17
29 7 5 7 9 51 7 13 8 55 42 9 81

Sample Output 2

Sample Input 3

10 400000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

题意:

给一个数列，求有多少个数对【l,r】使得,位置l->r的项的和为m的倍数。

解题思路：

对整个数列求前缀和，并对m取余数，那么当第l项和第r项mod m同余时，就是合乎要求的。

所以扫一遍r,使r分别等于1->n。用map记录每种余数对应的l的个数。

（不可能用数组来记录1E9种余数，一共只有1e5个数，最多只有1e5个值，用map把余数和个数对应起来）

当第r项的前缀和的余数为x时，那么与其配对的l的个数，即之前余数为x的个数。

且当位置i的前缀和可以被m除尽时，是可以以第一项为起点的。要在结果中补上。

思路来源：D - Candy Distribution（atcoder） (前缀和+取余技巧)

#include<stdio.h>
#include<map>
#define maxn 100005
using namespace std;
long long remainder[maxn];
long long a[maxn];
map<long long,long long>number_l;
int main()
{
    long long n,m,answer=0;
    scanf("%lld%lld",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        remainder[i]=remainder[i-1]+a[i];
        remainder[i]%=m;
        answer+=number_l[remainder[i]];
        number_l[remainder[i]]++;
    }
    answer+=number_l[0];/**0为起点自己为终点没有算进去，要补上**/
    printf("%lld\n",answer);
    return 0;
}