文章目录
- [[A - Buildings](https://atcoder.jp/contests/abc353/tasks/abc353_a)](#A - Buildings)
- [[B - AtCoder Amusement Park](https://atcoder.jp/contests/abc353/tasks/abc353_b)](#B - AtCoder Amusement Park)
- [[C - Sigma Problem](https://atcoder.jp/contests/abc353/tasks/abc353_c)](#C - Sigma Problem)
- [[D - Another Sigma Problem](https://atcoder.jp/contests/abc353/tasks/abc353_d)](#D - Another Sigma Problem)
- [[E - Yet Another Sigma Problem](https://atcoder.jp/contests/abc353/tasks/abc353_)](#E - Yet Another Sigma Problem)
A - Buildings
题目:
给一个数组H,从左往右找比第一个元素大的元素下标
思路:
模拟
java
static void solve() throws IOException {
int n = pInt();
int[] a = pIntArray(1);
int t = a[1];
for (int i = 2; i < a.length; i ++) {
if (a[i] > t) {
out.println(i);
return;
}
}
out.println(-1);
}
B - AtCoder Amusement Park
题目:
有一个最多容纳K个人的演出,有N组人在排队,第i组人有A~i~ 个人,要求每次放一组人进去且总人数不能超过K个人,当无法继续放入时开始演出,接着继续放人进入,直到没人排队为止,求开始演出的次数
思路:
模拟
java
static void solve() throws IOException {
int[] ins = pIntArray(0);
int n = ins[0], k = ins[1];
int[] a = pIntArray(0);
int ans = 0, cur = 0;
for (int i = 0 ; i < n; i ++) {
if (cur + a[i] > k) {
cur = 0;
ans ++;
}
cur += a[i];
}
if (cur > 0) ans ++;
out.println(ans);
}
C - Sigma Problem
题目:
定义 f ( x , y ) = ( x + y ) m o d 1 0 8 f(x,y) = (x+y) \mod 10^8 f(x,y)=(x+y)mod108 ,给一个正整数序列A,求 ∑ i = 1 N − 1 ∑ j = i + 1 N f ( A i , A j ) \sum^{N-1}{i=1} \sum^{N}{j=i+1}f(A_i, A_j) ∑i=1N−1∑j=i+1Nf(Ai,Aj)
思路:
观察发现A的顺序任意交换不会影响最后的结果,而f相当于x+y大于等于1e8的时候就会减去1e8,题目的数据A~i~<1e8就意味着x+y<2e8,最多减去一次1e8;
先进行排序,对于每个元素x,通过二分找到使得x+y>=1e8的第一个y,此时y右边的所有数加上x都会大于等于1e8,那么就需要减去这些元素的个数*1e8
java
static void solve() throws IOException {
final long mod = (long) 1e8;
int n = pInt();
int[] a = pIntArray(0);
long ans = 0;
Arrays.sort(a);
for (int i = 0; i < n; i ++) {
ans += (long) a[i] * (n - 1);
int l = i + 1, r = n - 1;
while (l < r) {
int mid = l + r >> 1;
if (a[i] + a[mid] >= mod) {
r = mid;
} else {
l = mid + 1;
}
}
if (l < n && a[i] + a[l] >= mod) {
ans -= (n - l) * mod;
}
}
out.println(ans);
}
D - Another Sigma Problem
题目:
给一个整数序列A,定义 f ( x , y ) f(x,y) f(x,y) 为将x和y前后拼接成的整数,求 ∑ i = 1 N − 1 ∑ j = i + 1 N f ( A i , A j ) \sum^{N-1}{i=1} \sum^{N}{j=i+1}f(A_i, A_j) ∑i=1N−1∑j=i+1Nf(Ai,Aj),结果对998244353取模
思路:
仔细观察其实可以发现,每个数x作为前面一部分的时候,相当于x*10^(y的位数)+y;
如何统计10^(y的位数)这一部分呢,二重循环是会超时的;但是可以从最后一个开始统计;
java
static void solve() throws IOException {
final long mod = 998244353;
int n = pInt();
long[] a = pLongArray(0);
long ans = 0;
long count = 0, sum = 0;
for (int i = n - 1; i >= 0; i --) {
ans = (ans + a[i] * count % mod + sum % mod) % mod;
long t = a[i], cnt = 0;
while (t > 0) {
cnt ++;
t /= 10;
}
sum += a[i];
count += (long) Math.pow(10, cnt);
}
out.println(ans % mod);
}
E - Yet Another Sigma Problem
题目:
定义 f ( x , y ) f(x,y) f(x,y)为两个字符串x和y的最长公共前缀的长度 ,给一个字符串数组S,求 ∑ i = 1 N − 1 ∑ j = i + 1 N f ( S i , S j ) \sum^{N-1}{i=1} \sum^{N}{j=i+1}f(S_i, S_j) ∑i=1N−1∑j=i+1Nf(Si,Sj)
思路:
用字典树实现记录前缀,也就是每个点都多少个字符串经过,最后统计这些点构成的公共前缀对答案的贡献度即可;
比如ad
、 abc
和 abd
,a
点经过3次,b
点经过2次,分别的贡献度为3和1
java
static Map<Integer, Integer> map = new HashMap<>();
static class Trie {
int[][] p = new int[26][(int) (3e5 + 5)];
int idx = 0;
{
for (int i = 0; i < 26; i ++) {
Arrays.fill(p[i], -1);
}
}
void insert(String s) {
int n = s.length();
int cur = 0;
for (int i = 0; i < n; i ++) {
int t = s.charAt(i) - 'a';
if (p[t][cur] == -1) {
p[t][cur] = ++ idx;
}
cur = p[t][cur];
map.compute(cur, (k, v) -> v == null ? 1 : v + 1);
}
}
}
static void solve() throws IOException {
int n = pInt();
String[] ins = pStringArray();
Trie root = new Trie();
long ans = 0;
for (int i = 0; i < n; i ++) {
root.insert(ins[i]);
}
for (int v : map.values()) {
ans += (long) v * (v - 1) / 2;
}
out.println(ans);
}