Mathematics + Thinking] ZZULIOJ 1531: a small interval L sum

Title Description

In a given sequence of integers, a small L wanted to find a continuous range, and this range can be divisible by k, would you please help small L count to meet the conditions of the longest interval length is.

Entry

The first line of the input two integers n, k. (1 <= n <= 105,1 <= k <100)
the next line of input integer n, the number of sequences in FIG.

Export

Output an integer, satisfying the condition longest length section, and if not, outputs 0

Sample input

5 7
1 2 4 1 1

Sample Output

3 

Solution:

Prefix and an array of arrays built up first. K mold during construction, and so in this case the array of all the number of prefixes belonging to [0, K-1], if there are two positions i, j, and the prefix array pre [i] == pre [j ], then do not K and prefix die array pre [j] -pre [i] result must be a multiple of K, ji is the length. 
So we aim to become a constantly looking for: pre [i] == pre [ j], save length = ji and updates. The following code is used: map is first initialized to -1, then encountered during traversal of the recording position on pre [i] of the first occurrence of

#include <bits / STDC ++ H.>
 the using  namespace STD; 
typedef Long  Long LL;
 const  int NN = 1E5; 

int N, K;
 int A [NN + 10 ]; // source array 
int pre [NN + 10 ]; // prefix and array 
Map < int , int > MP; // MP [n] = i represents the prefix and the number n in the array occurs first at position i 
int main () { 
    CIN >> N >> K;
     for ( int i = . 1 ; I <= N; I ++) { // read data
        Scanf ( " % D " , & A [I]); 
    }     
    for ( int I = . 1 ; I <= N; I ++) { // Construction prefix and the array and modulo K 
        pre [I] = (A [I] + pre [I- . 1 ])% K; 
        MP [pre [I]] = - . 1 ; // initialized to -1 
    } 
    int the maxlen = 0 ; // longest length 
    for ( int I = . 1 ; I <= N ; I ++ ) {
         IF (MP [pre [I]] == - . 1 ) {   
            MP [pre [I]] = I; // position into the first occurrence
        } The else { 
            the maxlen = max (the maxlen, I-mp [pre [i]]); // I is pre [i] At this time, a position mp [pre [i]] is the location of pre [i] of the first time , two subtraction of multiples k 
        } 
    } 
    COUT << the maxlen << endl;
     return  0 ; 
}

Another writing position with the vector, vector and array elements stored prefixes appear

#include <bits / STDC ++ H.>
 the using  namespace STD; 
typedef Long  Long LL;
 const  int NN = 1E5; 

int N, K;
 int A [NN + 10 ]; // source array 
int pre [NN + 10 ]; // prefix and an array of 
Vector < int > VE [ 110 ];
 int main () { 
    CIN >> N >> K;
     for ( int I = . 1 ; I <= N; I ++) { // read data 
        Scanf ( " % D",&a[i]);
    }    
    ve[0].push_back(0);
    for(int i = 1;i<=N;i++){ //构建前缀和数组并取余K 
        pre[i] = (a[i]+pre[i-1])%K;
        ve[pre[i]].push_back(i);
    }
    int maxlen = 0;
    for(int i = 0;i<K;i++){
        if(ve[i].size()>1)
            maxlen = max(maxlen,ve[i][ve[i].size()-1] - ve[i][0]); // last occurrence position - the position of the first occurrence 
    } 
    COUT << the maxlen << endl;
     return  0 ; 
}

Mathematics + Thinking] ZZULIOJ 1531: a small interval L sum

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