Codeforces 892 B. Wrath （递推）

Codeforces 892 B. Wrath （递推）

Description

Hands that shed innocent blood!There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.You are given lengths of the claws. You need to find the total number of alive people after the bell rings.

Input

The first line contains one integer n (1 ≤ n ≤ 10^6) — the number of guilty people.Second line contains n space-separated integers L1, L2, …, Ln (0 ≤ Li ≤ 10^9), where Li is the length of the i-th person’s claw.

Output

Print one integer — the total number of alive people after the bell rings.

Examples input

40 1 0 10

Examples output

1

题意

每个人都有一个长度为 li

思路

看完题意，我的心情是这样的：

显然，最右侧的人一定是可以存活下来的。

我们维护一个 cnt

若 cnt>0 说明当前位置在别人的攻击范围内，否则 ans+1更新 cnt 为 max(cnt−1,ai) 看对于 i

时间复杂度 O(n)

AC 代码

#include#define IO ios::sync_with_stdio(false);\ cin.tie(0);\ cout.tie(0);using namespace std;typedef __int64 LL;const int maxn = 2e6+10;LL a[maxn],n;void solve(){ LL cnt = a[n-1],ans = 1; for(int i=n-2; i>=0; i--) { if(!cnt)ans++; cnt = max(cnt-1,a[i]); } cout<>n; for(int i=0; i>a[i]; solve(); return 0;}

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