Codeforces 776C - Molly's Chemicals(思维+前缀和)

【Codeforces 776C - Molly's Chemicals(思维+前缀和)】的更多相关文章

Codeforces 776C - Molly's Chemicals(思维+前缀和)

题目大意:给出n个数(a1.....an),和一个数k,问有多少个区间的和等于k的幂 (1 ≤ n ≤ 10^5, 1 ≤ |k| ≤ 10, - 10^9 ≤ ai ≤ 10^9) 解题思路:首先,可以从题目得出k的幂最大不能超过n*ai=1e14.然后,我们先求出前缀和sum[1]...sum[n],k的i次幂为power[i].我们要求出一个区间和等于power[i]即sum[r]-sum[l]=power[i],转换一下sum[r]=sum[l]+power[i]. 所以我们只需要从1到…

codeforces 776C Molly's Chemicals(连续子序列和为k的次方的个数)

题目链接题意:给出一个有n个数的序列,还有一个k,问在这个序列中有多少个子序列使得sum[l, r] = k^0,1,2,3…… 思路:sum[l, r] = k ^ t, 前缀和sum[r] = sum[l-1] + k^t.即如果后面有一段序列使得sum[l,r] = k^t,那么它可以转化为前缀和相减使得这一段大小为k^t,即sum[i] = sum[j] + k^t (1 <= j < i). 那么对于处理好的每一个前缀和,直接往后面加上k^t(0<=t<=x,k^x是不…

CodeForces - 776C（前缀和+思维）

链接:CodeForces - 776C 题意:给出数组 a[n] ,问有多少个区间和等于 k^x(x >= 0). 题解:求前缀和,标记每个和的个数.对每一个数都遍历到1e5,记录到答案. #include <bits/stdc++.h> using namespace std; long long n, k; map<long long, long long> mp; int main() { scanf("%lld%lld", &n, &am…

C. Molly's Chemicals

题目链接:http://codeforces.com/problemset/problem/776/C C. Molly's Chemicals time limit per test 2.5 seconds memory limit per test 512 megabytes input standard input output standard output Molly Hooper has n different kinds of chemicals arranged in a lin…

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C. Molly's Chemicals time limit per test 2.5 seconds memory limit per test 512 megabytes input standard input output standard output Molly Hooper has n different kinds of chemicals arranged in a line. Each of the chemicals has an affection value, The…

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