codeforces 776C

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本文详细解析 CodeForces 平台上的 776C 题目“Molly's Chemicals”。该题要求找出序列中子序列的和等于特定数 k 的任意次幂的数量。文章提供了算法思路，包括使用前缀和及哈希表来高效解决问题，并附带了 AC 代码。

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codeforces 776C

题目：
C. 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].

C题，题意给你一个序列，让你找出子序列的和等于k的任意次幂有多少个.

遍历一次求出前缀和，然后减去用k^i次幂，再在map里寻找，找到则ans+=map【x】
注意处理1和-1；

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
__int64 a[100005];

int main()
{
    int n;int k;
    while(scanf("%d %d",&n,&k)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%I64d",&a[i]);
        }
        map<__int64,int> s;
        s[0]++;
        __int64 sum=0,ans=0;
        for(int i=0;i<n;i++){
            sum+=a[i];
            for(__int64 j=1;abs(j)<=1e17;j*=k){
                __int64 x=sum-j;
                if(s.count(x)) {ans+=s[x];}
                if(k==1||j==-1) break;
            }
            s[sum]++;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}