【codeforces 776C】Molly's Chemicals

【题目链接】:http://codeforces.com/contest/776/problem/C

【题意】



让你找区间[i,j]

使得sum[i..j]=k^t,这里t=0,1,2,3..

-10<=k<=10

求出区间个数;

【题解】

/*
    k^0==1
    要求选[l,r]
    sum[l..r]==k^t,t>=0,t=1,2,3...
    前缀和
    pre[i]-pre[j] =k^t; j∈[1..i-1]
    pre[i]=k^t+pre[j-1];
    把dic[k^t+pre[j-1]]++;这里t∈....
    累加前缀和之后直接ans+=dic[pre[i]];
    注意处理一下k=1和k=-1的情况(不一样的!);
*/

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)

typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 1e5+100;

LL pre[N],a[N],temp[200],ans = 0;
int n, k,ma = 0;
map <LL, LL> dic;

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    rei(n), rei(k);
    LL t = 1;
    temp[0] = 1;
    if (k == 1||k==-1)
    {
        if (k == 1)
        {
            ma = 0;
            temp[0] = 1;
        }
        else
        {
            ma = 1;
            temp[0] = 1;
            temp[1] = -1;
        }
    }
    else
    {
        while (abs(t*k) < 1e15)
        {
            ma++;
            t = t*k;
            temp[ma] = t;
        }
    }
    rep1(i, 1, n)
        rel(a[i]);
    rep1(i, 0, ma)
        dic[temp[i] + 0]++;
    rep1(i, 1, n)
    {
        pre[i] = pre[i - 1] + a[i];
        ans += dic[pre[i]];
        rep1(j, 0, ma)
            dic[pre[i] + temp[j]]++;
    }
    printf("%lld\n", ans);
    return 0;
}