D - Candy Distribution（atcoder） (前缀和+取余技巧)
D - Candy Distribution
Time limit : 2sec / Memory limit : 1024MB
Score : 400 points
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≤10^5
- 2≤M≤10^9
- 1≤Ai≤10^9
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
Copy
3 2
4 1 5
Sample Output 1
Copy
3
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
6Sample Input 3
10 400000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Sample Output 3
25
题意：给定n个数，求有多少对区间【i，j】和是m的倍数。
首先，前缀和sum[i]表示[1,i]的区间和，那么[L,R]的区间和即为sum[R]-sum[L-1].已知(sum[R]-sum[L-1])%m=0,得到sum[R]%m=sum[L-1]%m。当我们遍历到R时，只需要知道sum[R]%m的出现次数就行了，同时更新cnt[sum[R]%m]++。注意，如果a[i]%m=0,那么这种情况是有疏漏的，需要最后输出的时候加上cnt[0]。
#include
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll n,m,ans=0;
mapcnt;
ll sum[maxn],num[maxn];
int main()
{
scanf("%d%d",&n,&m);
sum[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%lld",&num[i]);
sum[i]=(sum[i-1]+num[i])%m;
ans+=cnt[sum[i]];
cnt[sum[i]]++;
}
printf("%lld\n",ans+cnt[0]);
}