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看到这道题首先就想到不停地遍历，找到符合条件的就可以计数
#include <bits/stdc++.h> using namespace std; int N; int a[100000]; int K; int
ans; int main() { cin>>N>>K; for(int i=0;i<N;i++) { cin>>a[i]; } for(int
i=0;i<N;i++) { if(a[i]%K==0) { ans++; } for(int j=i+1;j<N;j++) {
a[i]=a[i]+a[j]; if(a[i]%K==0) { ans++; } } } cout<<ans; return 0; }
当然，通过分析，两层for循环时间复杂度为O（n*n），运行超时
所以，只能想办法找到更好的办法
正确代码：（用到排列组合知识）
#include <bits/stdc++.h> using namespace std; long long
mod[100010]={0},add[100010]={0}; long long sum=0,a[100010]; int main() { int
n,k; long long cnt=0; cin>>n>>k; for(int i=1;i<=n;i++) { cin>>a[i]; } for(int
i=1;i<=n;i++) { sum=sum+a[i]; mod[i]=sum%k; add[mod[i]]++; } for(int
i=0;i<n;i++) { cnt+=add[i]*(add[i]-1)/2; } cout<<cnt+add[0]; return 0; }
