Codeforces_776_C_(思维)(前缀和)

C. Molly's Chemicals

time limit per test

2.5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

Note

Do keep in mind that k0 = 1.

In the first sample, Molly can get following different affection values:

• 2: segments [1, 1], [2, 2], [3, 3], [4, 4];
• 4: segments [1, 2], [2, 3], [3, 4];
• 6: segments [1, 3], [2, 4];
• 8: segments [1, 4].

Out of these, 2, 4 and 8 are powers of k = 2. Therefore, the answer is 8.

In the second sample, Molly can choose segments [1, 2], [3, 3], [3, 4].

题意：给定n个数字和一个k，问将任意相邻的一段数字加起来是k的幂的方法数。

一直没有思路，只能想到枚举起点终点。。。

看了题解，用的方法感觉挺巧妙的。

思路：对n个数求前缀和，记录下前缀和的种数(即前缀中能得到值s的种数)，然后对每个k的幂(power)求sum-power的和。

在其前缀中有多少种方式组成sum-power，那么当前就能多少种power。

#include <cstdio>
#include <cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define N 100005
#define LL long long

int num[N];
map<LL,int> ma;
int main()
{
    int n,k;
    LL tmp=0;
    scanf("%d%d",&n,&k);
    for(int i=0; i<n; i++)
        scanf("%d",&num[i]);
    LL power=1,sum=0,res=0;
    while(abs(power)<1e15)
    {
        ma.clear();
        ma[0]=1;
        sum=0;
        for(int i=0; i<n; i++)
        {
            sum+=num[i];
            res+=ma[sum-power];
            ma[sum]++;
        }
        power*=k;
        if(power==1)
            break;
        //cout<<res<<endl;
    }
    printf("%I64d
",res);
    return 0;
}