codeforces 776C

题意：给定n个元素的数组，找出某个区间所有元素之和为k的非负次幂，计算一共有几个符合条件的区间。

思路：这种题总是忍不住想去暴力遍历所有区间。没想到还有遍历k。sum[I] - sum[j] = k^n (n=0,1,2...) ,(I>=j)sum为前缀和。

枚举k^n (n=1,2,3..)因为n最多1e5,a[I] 最多1e9 .全部和加起来1e15 ，所以k^n 0~1e15 

#include<cstdio>#include<queue>#include<iostream>#include<vector>#include<map>#include<cstring>#include<string>#include<set>#include<stack>#include<algorithm>#define cle(a) memset(a,0,sizeof(a))#define inf(a) memset(a,ox3f,sizeof(a))#define ll long long#define FOR(i,n) for(int i=0;i<n;i++)using namespace std;const int INF = ( 2e9 ) + 2;const int maxn = 1e5+10;ll a[maxn];map<ll,int> mp;int main(){int n,k;scanf("%d%d",&n,&k);for(int i=1;i<=n;i++)scanf("%I64d",&a[i]);ll p=1,ans=0;while(abs(p)<1e15+10){ll sum=0;mp.clear();mp[0]=1;for(int i=1;i<=n;i++){sum+=a[i];ans+=mp[sum-p];mp[sum]++;}p*=k;if(p==1)break;}printf("%I64d\n",ans);}