Codeforces 892 B. Wrath (递推)
2017-11-18 11:45
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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
4 0 1 0 10
Examples output
1
题意
每个人都有一个长度为 li 的武器,相邻的两个人之间距离为 1 ,同一时间所有人使用武器攻击左边的人,问最后存活下来的人数。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 代表右侧延伸到当前位置的武器长度,
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 来说是否可以攻击到更远的位置。
时间复杂度 O(n) 。
AC 代码
#include<bits/stdc++.h> #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<<ans<<endl; } int main() { IO; cin>>n; for(int i=0; i<n; i++) cin>>a[i]; solve(); return 0; }
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