ICM Technex 2017 and 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：给你一个n和k然后再给你n个数，看从这n个数中可以去多少个区间使得区间的和等于k的j次方（j=0,1,2,3,4，，，，）；

思路：首先sum[i]表示前i个数的和，我们需要找的是sum[i]-sum[j]=k^(0,1,2,3,,),根据这个情况在一一枚举即可求出。

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 1e5 + 20;
map<LL, int>mipos;
map<LL, int>num;
const LL uppos = 1e14;
const LL upnag = -1e14;
LL sum;
LL mypow(LL k, LL b)
{
LL ans = 1;
for (int i = 1; i <= b; ++i)
{
ans *= k;
}
return ans;
}
void work()
{
int n;
LL k;
cin >> n >> k;
num[0] = 1;
LL ans = 0;
for (int i = 1; i <= n; ++i)
{
int x;
cin >> x;
sum += x;
int sel = 0;
for (int j = 0; ; ++j)
{
LL now = mypow(k, j);
if (now > uppos) break;
if (now < upnag) break;
ans += num[sum - now];
sel++;
if (k == 1) break;
if (k == -1 && sel == 2) break;
}
num[sum]++;
}
cout << ans << endl;
}
int main()
{
work();
return 0;
}