牛客多校第三场 G Removing Stones(分治+线段树)
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牛客多校第三场 G Removing Stones(分治+线段树)
题意:
给你n个数,问你有多少个长度不小于2的连续子序列,使得其中最大元素不大于所有元素和的一半
题解:
分治+线段树
线段树维护最大值的位置,每次分治找就找这个最大值,然后看最大值在哪个序列是可行的
怎么看最大值所在的序列是否可行呢?
我们用一个前缀和维护区间和
\[
max<=\frac{1}{2}(sum[r]-sum[l])\\
2*max-(sum[r]-sum[l])<=0\\
\]
这个最大值在这一段区间内都有可能出现
为了得到总的方案数
我们在当前区间内枚举靠近最大值出现的位置pos的那一段,然后二分右\左端点,即可得到一段最小的区间[i,t]或者[t,i]可以满足 区间内最大元素不大于所有元素和的一半的条件,然后计数就加上容斥后的r-t+1或者t-l+1这一段即可
代码:
/**
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* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
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* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
pii Max[maxn << 2];
LL sum[maxn];
int a[maxn];
void push_up(int rt) {
Max[rt] = max(Max[ls], Max[rs]);
}
void build(int l, int r, int rt) {
if(l == r) {
Max[rt] = make_pair(a[l], l);
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
push_up(rt);
}
pii query(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
return Max[rt];
}
int mid = (l + r) >> 1;
pii ans = make_pair(0, 0);
if(L <= mid) ans = max(ans, query(L, R, lson));
if(R > mid) ans = max(ans, query(L, R, rson));
return ans;
}
//二分最短区间大于后缀
int find1(int l, int r, LL x) {
int L = l, R = r + 1, res = l - 1;
while (L <= R) {
int mid = (L + R) >> 1;
if (sum[r] - sum[mid - 1] >= x) {
L = mid + 1;
res = mid;
} else R = mid - 1;
}
return res;
}
int find2(int l, int r, LL x) {
int L = l - 1, R = r, res = r + 1;
while (L <= R) {
int mid = (L + R) >> 1;
if (sum[mid] - sum[l - 1] >= x) {
R = mid - 1;
res = mid;
} else L = mid + 1;
}
return res;
}
LL ans = 0;
int n;
void solve(int l, int r) {
if(l == r) return;
int mid = query(l, r, 1, n, 1).second;
if(l < mid) solve(l, mid - 1);
if(r > mid) solve(mid + 1, r);
if(mid - l < r - mid) {
for(int i = l; i <= mid; i++) {
int t = find2(mid + 1, r, 2 * a[mid] - (sum[mid] - sum[i - 1]));
ans += r - t + 1;
}
} else {
for(int i = mid; i <= r; i++) {
int t = find1(l, mid - 1, 2 * a[mid] - (sum[i] - sum[mid - 1]));
ans += t - l + 1;
}
}
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int T;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
sum[0] = 0;
ans = 0;
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
sum[i] = sum[i - 1] + a[i];
}
build(1, n, 1);
solve(1, n);
printf("%lld\n", ans);
}
return 0;
}
标签:Stones,int,Removing,最大值,ans,mid,res,include,第三场 来源: https://www.cnblogs.com/buerdepepeqi/p/11324366.html