Codeforces 776C Molly's Chemicals (思维)

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原题地址:http://codeforces.com/contest/776/problem/C

题意:：给定$n$个数，问和等于$k$的次方的连续区间有多少个。

思路:首先最容易想到的就是枚举区间范围,但是这题数据范围1e5,那么$O(n^2)$的复杂度显然不可行.那么可以转化一下式子,设现在区间的值是由前缀和得到的.所以满足题木要求就是满足下面的式子$sum[j]=k^x+sum[i],(i<=j)$,也就是$sum[i]=sum[j]-k^x$,那么我们只需要预处理$k^x$,然后枚举$j$,这样子复杂度就降下来了.

#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
int n, k;
ll a[maxn];
ll sum[maxn];
map<ll, int>mp;

int main() {
    ll MAX = 0;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
        scanf("%I64d", &a[i]);
        MAX += abs(a[i]);
    }
    ll p = 1;
    ll ans = 0;
    while (abs(p) <= MAX) {
        mp.clear();//每次都清空是因为要求j>=i,不然会多算
        mp[0] = 1;
        ll res = 0;
        for (int i = 1; i <= n; ++i) {
            res += a[i];
            ans += mp[res - p];
            mp[res]++;
        }
        p *= k;
        if (p == 1) break;
    }
    printf("%I64d\n", ans);
    return 0;
}