codeforces 766C 数学

Molly’s Chemicals

time limit per test2.5 seconds

memory limit per test512 megabytes

inputstandard input

outputstandard 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].

题目链接

题意：给你一个有n个数的集合和一个k，让你求有多少个连续的子集，里面的数值全部加起来为k的幂。

解题思路：不难想到如果要求[a,b]内所有数的和的最简单的方法就是[1,b]-[1,a]，那因此我们可以先求出所有前1-n项的和，然后因为最大值应该为1e14 (1e9 * 1e5)，即使是2的次幂也最多为50个，其中，1和-1单独考虑，每次只要知道[a,b]的和为k的非负整数幂函数的其中一个就好。

所以我们可以用map记录前1-n项出现过的和和它的个数，最后在每输入一个数时遍历k的非负整数次幂即可。代码如下：

#include<cstdio>
#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll inf=100000000000000LL;
map<ll,ll>p;
ll n,k,a[100],pl,b,s,ans;
int main(){
scanf("%lld%lld",&n,&k);
a[0]=1;
for(pl=1;pl<=64;pl++){
a[pl]=a[pl-1]*k;
if(abs(a[pl])>=inf) break;
}
if(k==1)    pl=0;
else if(k==-1)  pl=1;
b=0;
p[0]++;
for(ll i=1;i<=n;i++){
scanf("%lld",&s);
b=b+s;
for(ll j=0;j<=pl;j++){
ll aa=b-a[j];
ans+=p[aa];
}
p[b]++;
}
printf("%lld\n",ans);
return 0;
}