abc353

C - Sigma Problem

Problem Statement

For positive integers 𝑥x and 𝑦y, define 𝑓(𝑥,𝑦)f(x,y) as the remainder of (𝑥+𝑦)(x+y) divided by 108108.

You are given a sequence of positive integers 𝐴=(𝐴1,…,𝐴𝑁)A=(A1​,…,AN​) of length 𝑁N. Find the value of the following expression:

∑𝑖=1𝑁−1∑𝑗=𝑖+1𝑁𝑓(𝐴𝑖,𝐴𝑗)i=1∑N−1​j=i+1∑N​f(Ai​,Aj​).

Constraints

• 2≤𝑁≤3×10^5 2≤N≤3×10^5
• 1≤𝐴𝑖<1081≤Ai​<108
• All input values are integers.

Sample Input 1

3
3 50000001 50000002

Sample Output 1

100000012

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题意：数组里的数两两相加对1e8取模后求和。

数组大小3e5暴力肯定超时。每个数都会加n-1次，所以先将这些数加起来，然后减掉需要取模的那些和的1e8就行。所以先排序方便使用lowerbound判断有几个和大于1e8，减去相应数量的1e8就行。

（因为1e8不是longlong wa了……）

总而言之看代码。

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;
typedef pair<LL, LL> PII;
const double pi = 3.1415926;
const int N = 1000010, INF = 1e8;
LL n, m;
LL a[N];

int main() {
    cin >> n; 
    LL ans = 0, t;
    for(int i = 0;i < n;i ++) cin >> a[i], ans += a[i] * (n - 1);
    sort(a, a + n);
    
    for(LL i = 0;i < n;i ++) {
        t = lower_bound(a, a + n, 100000000 - a[i]) - a;
        t = max(t, i + 1);
        if(t > i) ans -= (n - t) * (1e8);
    }

    cout << ans << '\n';

    return 0;
}

E - Yet Another Sigma Problem

Problem Statement

For strings 𝑥x and 𝑦y, define 𝑓(𝑥,𝑦)f(x,y) as follows:

• 𝑓(𝑥,𝑦)f(x,y) is the length of the longest common prefix of 𝑥x and 𝑦y.

You are given 𝑁N strings (𝑆1,…,𝑆𝑁)(S1​,…,SN​) consisting of lowercase English letters. Find the value of the following expression:

∑𝑖=1𝑁−1∑𝑗=𝑖+1𝑁𝑓(𝑆𝑖,𝑆𝑗)i=1∑N−1​j=i+1∑N​f(Si​,Sj​).

Constraints

• 2≤𝑁≤3×1052≤N≤3×105
• 𝑆𝑖Si​ is a string consisting of lowercase English letters.
• 1≤∣𝑆𝑖∣1≤∣Si​∣
• ∣𝑆1∣+∣𝑆2∣+…+∣𝑆𝑁∣≤3×105∣S1​∣+∣S2​∣+…+∣SN​∣≤3×105
• All input numbers are integers.

Input

The input is given from Standard Input in the following format:

Sample Input 1

3
ab abc arc

Sample Output 1

4

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题意：给一些小写字母组成的字符串，求他们每两个之间的公共前缀的长度之和。

字典树

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;
typedef pair<LL, LL> PII;
const double pi = 3.1415926;
const int N = 500010, INF = 998244353;
int n, m;
int tr[N][26], cnt[N], idx;
LL ans = 0;

void add(string s) {
    int p = 0;
    for(auto t : s) {
        int u = t - 'a';
        if(!tr[p][u]) tr[p][u] = ++ idx;
        p = tr[p][u];
        ans += cnt[p];
        cnt[p] ++;
    }
    
    
}

int main() {
    cin >> n; 
    
    for(int i = 0;i < n;i ++) {
        string s;
        cin >> s;
        add(s);
    }


    cout << ans << '\n';

    return 0;
}