Codeforces Round #400 (Div. 1 + Div. 2, combined)C. Molly's Chemicals

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：给你n个数和一个数k，问有多少个区间满足区间和等于k的x次方（x取值0到正无穷）。

思路：这种问区间和的问题无疑是要用前缀和的了，然后如果n^2暴力枚举区间肯定会超时的，此时需要扭转思路，枚举k的x次方，x取值最多到60左右就可以了，因为每个数范围是-1e9到1e9，故区间和的下限是-1e14上限是1e14。然后就没什么难度了。注意特判1和-1的情况，下面给代码：

#include<iostream>
#include<cmath>
#include<queue>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<utility>
#include<map>
#include<vector>
#define maxn 100005
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;
const double eps = 1e-5;
map<LL, int>m;
LL a[maxn], p[65];
int main(){
int n, k;
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++){
scanf("%lld", &a[i]);
a[i] += a[i - 1];
}
m[0] = 1;
LL ans = 0;
if (k == 1){
for (int i = 1; i <= n; i++){
ans += m[a[i] - 1];
m[a[i]]++;
}
}
else if (k == -1){
for (int i = 1; i <= n; i++){
ans += m[a[i] - 1];
ans += m[a[i] + 1];
m[a[i]]++;
}
}
else{
p[0] = 1;
int pos = 1;
while (1){
p[pos] = p[pos - 1] * k;
if (p[pos] > 1e14 || p[pos] < -1e14)
break;
pos++;
}
for (int i = 1; i <= n; i++){
for (int j = 0; j < pos; j++){
ans += m[a[i] - p[j]];
}
m[a[i]]++;
}
}
printf("%lld\n", ans);
}