Atcoder--Candy Distribution II--前缀和+map

题目描述

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

输入

Input is given from Standard Input in the following format:

N M
A1 A2 … AN

输出

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.

样例输入

3 2
4 1 5

样例输出

3

提示

The sum A_l+A_l+1+…+A_r 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.

题意：
求能够整除m的区间的个数
开始用了一遍前缀和，再用两个for循环遍历，然后T了
再看模数的取值范围，开不了数组，于是就用map来解决

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
#define N 1005
///int num[maxn];
///int sum[maxn];
map<ll,ll>mp;
int n;
ll m;
start{
    n=read,m=read;
    ll num,sum = 0,ans = 0;
        mp.clear();
        mp[0]++;
        for(int i=1; i<=n;i++) {
            num=read;
            sum+=num;
            sum%=m;
            ans+=mp[sum];
            mp[sum]++;
        }
        printf("%lld\n",ans);
    end;
}