思维技巧

description

Molly Hooper has n different kinds of chemicals arranged in a line. Each of the chemicals has an affection value, The i-th of them has affection value ai.
Molly wants Sherlock to fall in love with her. She intends to do this by mixing a contiguous segment of chemicals together to make a love potion with total affection value as a non-negative integer power of k. Total affection value of a continuous segment of chemicals is the sum of affection values of each chemical in that segment.
Help her to do so in finding the total number of such segments.

Input

The first line of input contains two integers, n and k, the number of chemicals and the number, such that the total affection value is a non-negative power of this number k. (1 ≤ n ≤ 105, 1 ≤ |k| ≤ 10).
Next line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — affection values of chemicals.

Output

Output a single integer — the number of valid segments.

Examples

input

4 2
2 2 2 2

output

8

input

4 -3
3 -6 -3 12

output

3

题解

首先输入n，k。之后输入n个数，求子序列和为 k ^ m ( m = 0,1,2,3 ...) 的数量。
定义 sum[i] 为前i个数的和，则有 sum[i] - sum[j] == k ^ m, 但是算两次 sum[i] 和 sum[j] 时间复杂度就较大；我们来转化一下公式，使它变成 以下这样 sum[i] - k ^ m == sum[j] ，我们可以发现，如果存在 sum[i] - k ^ m 的值，那么就可以说明有这么一对 i ， j 使结果存在，但是时间复杂度为 o( m lg n )。

code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <map>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <cctype>
#include <set>
#define For(i,a,b) for(int (i)=(a);(i) < (b); ++(i))
#define Fors(i,a,b) for(int (i)=(a);(i) > (b); --(i))
#define sd(x) cout << "start debug:" << (x) << endl;
#define ed(x) cout << "end debug:" << (x) << endl;
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long LL;
const LL INF = 1e14;

int n;
LL k,sum[100050];
map<LL,LL>x;
set<LL>a;
set<LL>::iterator it;

int main()
{
    while(~scanf("%d %lld",&n,&k))
    {
        x.clear();
        a.clear();
        sum[0]=0;
        for(int i=1;i<=n;i++) scanf("%lld",&sum[i]),sum[i]+=sum[i-1];

        a.insert(1);
        LL temp=k;
        for(int i=1;i<=60;i++){
            if(temp>INF) break;
            a.insert(temp);
            temp*=k;
        }

        LL ans=0;
        x[0]=1;
        for(int i=1;i<=n;i++){
            for(it=a.begin();it!=a.end();it++) ans+=x[sum[i]-*it];
            x[sum[i]]++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}