Looksery Cup 2015 F - Yura and Developers 单调栈+启发式合并

F - Yura and Developers

第一次知道单调栈搞出来的区间也能启发式合并。。。

你把它想想成一个树的形式，可以发现确实可以启发式合并。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std;const int N = 3e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;int n, k, tot, a[N], stk[N], L[N], R[N], sum[N];
vector<int> num[1000007];void add(int &a, int b) {a += b; if(a >= mod) a -= mod;
}
int cal(int x, int l, int r) {auto it1 = upper_bound(num[x].begin(), num[x].end(), r);auto it2 = lower_bound(num[x].begin(), num[x].end(), l);return it1 - it2;
}
int main() {scanf("%d%d", &n, &k);num[0].push_back(0);for(int i = 1; i <= n; i++) {scanf("%d", &a[i]);sum[i] = (sum[i-1]+a[i]%k)%k;num[sum[i]].push_back(i);}a[0] = a[n+1] = inf;stk[++tot] = 0;for(int i = 1; i <= n; i++) {while(tot && a[stk[tot]] < a[i]) tot--;L[i] = stk[tot];stk[++tot] = i;}tot = 0;stk[++tot] = n+1;for(int i = n; i >= 1; i--) {while(tot && a[stk[tot]] <= a[i]) tot--;R[i] = stk[tot];stk[++tot] = i;}LL ans = 0;for(int i = 1; i <= n; i++) {if(i - L[i] < R[i] - i) {for(int j = L[i]; j < i; j++)ans += cal((sum[j]+a[i])%k, i, R[i]-1);} else {for(int j = i; j < R[i]; j++) {ans += cal((sum[j]-(a[i]%k)+k)%k, L[i], i-1);}}}printf("%lld\n", ans - n);return 0;
}/*
*/

转载于:https://www.cnblogs.com/CJLHY/p/9916153.html

Looksery Cup 2015 F - Yura and Developers 单调栈+启发式合并相关推荐

Looksery Cup 2015 F. Yura and Developers（单调栈+二分+分治）（难*）
题目链接题意:给定一个数组,问有多少区间满足:去掉最大值之后,和是k的倍数. 思路:日后补. #include<bits/stdc++.h> using namespace std; t ...
【Codeforces549F】Yura and Developers [单调栈][二分]
Yura and Developers Time Limit: 20 Sec Memory Limit: 512 MB Description Input Output Sample Input 4 ...
【Codeforces 549F】Yura and Developers | 单调栈、启发式合并、二分
题目链接:https://codeforces.com/problemset/problem/549/F 题目大意: 给定一个序列和一个mod值,定义[l,r]合法当l到r的所有元素和减去其中的最大值 ...
[CF 526 F] Pudding Monsters（单调栈 + 线段树）
CF526F Pudding Monsters problem solution code problem luogu翻译 solution observation :每行每列恰好有一个棋子,所以如果 ...
Looksery Cup 2015 H. Degenerate Matrix 数学
H. Degenerate Matrix Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/549/ ...
codeforces Looksery Cup 2015 H Degenerate Matrix
The determinant of a matrix 2 × 2 is defined as follows: A matrix is called degenerate if its determ ...
Codeforces - Yura and Developers
题目链接:Codeforces - Yura and Developers 有一个显然的2个log做法. 分治最大值,然后枚举小区间启发式计算答案. 但是其实可以一个log,因为分治的时候,对于区间形 ...
codeforces 549F Yura and Developers（分治、启发式合并）
codeforces 549F Yura and Developers 题意给定一个数组,问有多少区间满足:去掉最大值之后,和是k的倍数. 题解分治,对于一个区间,找出最大值之后,分成两个区间. ...
COCI 2015/2016 PROKLETNIK（单调栈+线段树）
COCI 2015/2016 PROKLETNIK 题目大意给定序列A1..NA_{1..N}A1..N,分别求出QQQ组询问区间[L,R][L,R][L,R]中最长的连续子序列Al..rA_{l ...

Looksery Cup 2015 F - Yura and Developers 单调栈+启发式合并

Looksery Cup 2015 F - Yura and Developers 单调栈+启发式合并相关推荐

最新文章

热门文章