AtCoder beginner contest 166 E

题目意思

给出n个数,\(s_1...s_i...s_n\)
然后让你求\(|j - i| = s[i]+s[j]\)出现的次数

solution

我们就直接假设j>i;
然后把上边的式子移一下项就得到了

\[j - s[j] = i+s[i] \]

因为这个题有2秒和1G
所以我们就直接开一个map存每个i+s[i] 出现的次数

然后看j-s[j]出现了几次
复杂度就是\(O(n)\)

code

/*
	Auther:_Destiny
	time:2020.5.3
*/
#include <bits/stdc++.h>
#define ll long long
#define N 200010
#define M 1010

using namespace std;
int n;
int f[N];
map<int, int> ss;

int read() {
	int s = 0, f = 0; char ch = getchar();
	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

int main() {
	n = read();
	for (int i = 1; i <= n; i++)
		f[i] = read();
	for (int i = 1; i <= n; i++) {
		int x = i + f[i];
		ss[x]++;
	}
	ll ans = 0;
	for (int i = 1; i <= n; i++)
		ans += ss[i - f[i]];
	cout << ans << "\n";
}