Codeforces Round #400 C. Molly's Chemicals [二分]

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Molly试图通过混合特定化学物质段落来制作爱情药水，使其总好感值为k的非负整数次幂。本篇介绍了一种二分查找法解决该问题的方法，包括预处理区间和枚举区间的起点。

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].

题解

二分解法

预处理区间，枚举区间起点，二分寻找终点
注意终点可能是多个

复杂度 $o(n\log n)$

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define MAX_N 100005
typedef long long LL;
using namespace std;
typedef pair<LL,int> P;
P q[MAX_N];

int n,k;

LL Z(LL x){
    return x<0?-x:x;
}
bool cmp(P x,P y){
    if(x.first==y.first) return x.second<y.second;
    return x.first<y.first;
}
LL inf=1LL<<60;

int find_lower(LL now,int i){
    int l=-1,r=n+1,mid;
    while(r-l>1) {
        mid=(l+r)>>1;
        if(q[mid].first>now+q[i].first||
        (q[mid].first==now+q[i].first&&q[mid].second>q[i].second)) r=mid;
        else l=mid;
    }
    return r;
}

int find_upper(LL now,int i){
    int l=0,r=n+1,mid;
    while(r-l>1) {
        mid=(l+r)>>1;
        if(q[mid].first>now+q[i].first) r=mid;
        else l=mid;
    }
    return l;
}

int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%I64d",&q[i].first);
    for(int i=1;i<=n;i++){
        q[i].first+=q[i-1].first;
        q[i].second=i;
    }
    q[0].second=0;
    q[0].first=0;

    sort(q,q+n+1,cmp);
    LL ans=0;
    for(LL now=1;Z(inf/now)>Z(k);now*=k){
        for(int i=0;i<=n;i++){
            int low=find_lower(now,i);
            int top=find_upper(now,i);
            if(q[low].first!=now+q[i].first) continue;
            ans+=top-low+1;
        }
        if(k==1||(k==-1&&now==-1)) break;
    }
    printf("%I64d",ans);
    return 0;
}