[Jizhong analog 2019.08.01] [] columns JZOJ2644

Meaning of the questions:

　　In a sequence of positive integers length of $ N $ $ \ {a_i \} $, the definition of a "legal" sub-range, and which can be divisible by $ K $. Seeking original sequence includes the number of "legal" sub-section?

　　A total of $ T $ data set.

　　$ 1 \ T \ 20, \ quad 1 \ N \ 5 * 10 ^ 4, \ quad 1 \ K \ 10 ^ 6, \ quad 1 \ the a_i \ 10 ^ 9 $

analysis:

　　At first glance unexpected $ O (n \, logn) $ approach to write about $ O (n) \ sim O (n ^ 2) $ violent it.

　　Here $ O (n) $ prefix and the pretreatment time, $ O (n ^ 2) $ is an enumeration endpoint time interval. Every time found a "legal" sub-section on $ ans ++ $.

　　Consider, we do this is based on the equation: $$ pre [r] -pre [l-1] = \ sum \ limits ^ {r} _ {i = l} a_i \ equiv 0 \; (mod \ , K) $$

　　Then adapt it about, clearly $$ pre [r] \ equiv pre [l-1] \; (mod \, K) $$

　　Dying disease in shock and sat up, this question open barrel I see the line! Interval any two locations prefixes and die with more than $ K $ as the endpoint is composed of "legal", so we opened the barrel of each mold $ K $ remainder are counted, the number of combinations using seek answers.

　　Time complexity O (K), the space complexity of O (N + K).

Implementation (100):

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define IL inline
using namespace std;
const int N=5e4;
const int K=1e6;
typedef long long LL;

    int n,T;
    LL k,a[N+3],s[N+3],c[N+3],ans;
    int cnt[K+3];

int main () {
    c[0]=c[1]=0;    c[2]=1;
    for(int i=3;i<=N;i++)
        c[i]=c[i-1]+i-1;

    scanf("%d",&T);
    while(T--){
        scanf("%lld%d",&k,&n);
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        
        memset(cnt,0,sizeof cnt);
        s[0]=0;    cnt[0]=1;
        for(int i=1;i<=n;i++){
            s[i]=(s[i-1]+a[i])%k;
            cnt[s[i]]++;
            
        }
        
        ans=0;
        for(int i=0;i<k;i++)
            years + = c [cnt [i]];
        
        printf("%lld\n",ans);
        
    }

    return 0;

}

View Code

summary:

　　For the count, we can usually put the principle of deformation equation, find the law, can be violent from $ ans ++ $ evolved.