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【小码匠自习室】AGC023-A ：为啥总是N连发？为啥总遇到大神？

2022-08-08 13:52:00 【小码匠】

碎碎念

  • 老码农惯用小伎俩，每学一个知识点，总是N连发刷题，美其名曰：要慢下来，慢下来，慢下来
  • 又遇到大神的代码，大神连刷2天，就为了执行时长从15ms -> 8ms，估计对性能有洁癖
    • AC执行最长时长：1807ms
    • 大神时长：8ms
  • 意外惊喜：tourist大神也有不会做的题目，这次比赛的E题，大神也没搞定，排名第三，真是罕见
    • 看大神的代码犹如沐浴春风，望尘莫及，简洁的令人生畏。

标签

  • 累积和、map

题目地址

  • A - Zero-Sum Ranges
    • https://atcoder.jp/contests/agc023/tasks/agc023_a

题目描述

有一个整数序列A，其长度为N。

求 A 的和为 0 的非空连续子序列的个数。注意我们是在统计取出子序列的方式。也就是说，即使有的两个子序列的内容相同，如果它们取自不同的位置，也当成不同序列.

约束条件

  • 1 \leq N \leq 2 \times 10^5
  • -10^9 \leq A_i \leq 10^9
  • 所有输入值都是整数.

输入

输入格式如下:

N
A1 A2 ...... AN

输出

求和为 0 的 A 的非空连续子序列的个数.

输入示例 1

6
1 3 -4 2 2 -2

输出示例 1

3

There are three contiguous subsequences whose sums are 0: (1,3,-4), (-4,2,2) and (2,-2).

输入示例 2

7
1 -1 1 -1 1 -1 1

输出示例 2

12

In this case, some subsequences that have the same contents but are taken from different positions are counted individually. For example, three occurrences of (1, -1) are counted.

输入示例 3

5
1 -2 3 -4 5

输出示例 3

0

There are no contiguous subsequences whose sums are 0.

题解

小码匠题解

  • 楼下是大神：tourist题解，代码的每一个细节都值得学习
void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n;
    cin >> n;
    unordered_map<long long , long long > b;
    vector<long long> a(n);
    cin >> a[0];
    b[0] = 0;
    b[a[0]]++;
    for(int i = 1; i < n; ++i) {
        cin >> a[i];
        a[i] += a[i - 1];
        b[a[i]]++;
    }
    long long ans = 0;
    for(int i = 0; i < n; ++i) {
        ans += b[a[i]];
        if(a[i] != 0) {
            ans--;
        }
        b[a[i]]--;
    }
    cout << ans;
}

参考题解（tourist）

void best_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    
    int n;
    cin >> n;
    
    map<long long, int> mp;
    mp[0] = 1;
    
    long long s = 0;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        s += x;
        ans += mp[s];
        mp[s]++;
    }
    cout << ans << '\n';
}

参考题解

#include <iostream>
#include <vector>
#include <map>
using namespace std;

int main() {
    int N; cin >> N;
    vector<long long> a(N);
    for (int i = 0; i < N; ++i) cin >> a[i];

    // 累積和と連想配列
    vector<long long> s(N+1, 0);
    map<long long, long long> nums;
    for (int i = 0; i < N; ++i) s[i+1] = s[i] + a[i];
    for (int i = 0; i < N+1; ++i) nums[s[i]]++;

    // 集計処理
    long long res = 0;
    for (auto it : nums) {
        long long num = it.second; // it.first が it.second 個ある
        res += num * (num - 1) / 2;
    }
    cout << res << endl;
}

又遇大神

#pragma GCC optimize ("O3")
#pragma GCC target ("avx2")
//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please
 
const int CM = 1 << 17, CL = 12;
char cn[CM + CL], * ci = cn + CM + CL, * owa = cn + CM, ct;
const ll ma0 = 1157442765409226768;
const ll ma1 = 1085102592571150095;
const ll ma2 = 71777214294589695;
const ll ma3 = 281470681808895;
const ll ma4 = 4294967295;
inline int getint() {
	if (ci - owa > 0) {
		memcpy(cn, owa, CL);
		ci -= CM;
		fread(cn + CL, 1, CM, stdin);
	}
	int pn = 1;
	if (*ci == '-') {
		pn = -pn;
		ci++;
	}
	ll tmp = *(ll*)ci;
	int dig = ((tmp & ma0) ^ ma0) ? 68 - __builtin_ctzll((tmp & ma0) ^ ma0) : 0;
	tmp = tmp << dig & ma1;
	tmp = tmp * 10 + (tmp >> 8) & ma2;
	tmp = tmp * 100 + (tmp >> 16) & ma3;
	tmp = tmp * 10000 + (tmp >> 32) & ma4;
	ci += (64 - dig >> 3);
	while ((ct = *ci++) >= '0') tmp = tmp * 10 + ct - '0';
	return pn * tmp;
}
 
ll A[200001];
void pakuri_sort(int N, ll A[]) {
	const int b = 8;
	ll tmp[200001];
	rep(k, 4) {
		int kazu[1 << b] = {}, kazu2[1 << b] = {};
		rep(i, N) kazu[A[i] >> k * b & ((1 << b) - 1)]++;
		rep(i, (1 << b) - 1) kazu[i + 1] += kazu[i];
		for (int i = N - 1; i >= 0; i--) tmp[--kazu[A[i] >> k * b & ((1 << b) - 1)]] = A[i];
		k++;
		rep(i, N) kazu2[tmp[i] >> k * b & ((1 << b) - 1)]++;
		rep(i, (1 << b) - 1) kazu2[i + 1] += kazu2[i];
		for (int i = N - 1; i >= 0; i--) A[--kazu2[tmp[i] >> k * b & ((1 << b) - 1)]] = tmp[i];
	}
}
 
int main() {
	//cin.tie(0);
	//ios::sync_with_stdio(false);
 
 
	int N = getint();
	rep1(i, N) {
		A[i] = getint() + A[i - 1];
	}
 
	pakuri_sort(N + 1, A);
 
	ll kotae = 0;
	ll mae = 1e18;
	int kazu = 0;
	rep(i, N + 1) {
		if (mae != A[i]) {
			mae = A[i];
			kazu = 0;
		}
		kotae += kazu++;
	}
 
	printf("%lld\n", kotae);
 
	Would you please return 0;
}
原网站

版权声明
本文为[小码匠]所创，转载请带上原文链接，感谢
https://cloud.tencent.com/developer/article/2068169

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