C - Lamps

标签：个数   namespace   thml   long   integer   memset   element   turn   cin   

We have $N$ bulbs arranged on a number line, numbered $1$ to $N$ from left to right. Bulb $i$ is at coordinate $i$.

Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity $d$ at coordinate $x$, the bulb illuminates the segment from coordinate $\mathrm{x-d-0.5}$ to $\mathrm{x+d+0.5}$. Initially, the intensity of Bulb $i$ is ${\mathrm{Ai}}^{}$. We will now do the following operation $K$ times in a row:

• For each integer $i$ between $1$ and $N$ (inclusive), let ${\mathrm{Bi}}^{}$ be the number of bulbs illuminating coordinate $i$. Then, change the intensity of each bulb $i$ to ${\mathrm{Bi}}^{}$.

Find the intensity of each bulb after the $K$ operations.

题意：坐标轴上1-n有n个灯泡，给出每个灯泡的初始亮度，进行k次操作，每次用照亮该坐标的灯泡个数更新该坐标上灯泡的亮度，最后输出每个灯泡亮度。

思路：常规做法一定会超时，所以用差分数组+每个灯泡亮度最大时退出循环特判。

#include <iostream>
#include <string.h>
using namespace std;
#define ll long long
const int N=2e5+5;
int a[N],b[N];

int main(){
    int n,k;cin>>n>>k;
    for(int i=0;i<n;i++)
        cin>>a[i];
    while(k--){
    int flag=0;
    memset(b,0,sizeof(b));
    for(int i=0;i<n;i++){
        b[max(i-a[i],0)]++;
        if(i+a[i]+1<=n-1)
        b[i+a[i]+1]--;
    }
    a[0]=b[0];
    for(int i=1;i<n;i++){
        b[i]+=b[i-1];
        a[i]=b[i];
    }
    for(int i=1;i<n;i++){
        if(a[i]!=n){
            flag=1;
            break;
        }
    }
    if(flag==0) break;
    }
    for(int i=0;i<n;i++)
        cout<<a[i]<<‘ ‘;
    return 0;
}

C - Lamps

标签：个数   namespace   thml   long   integer   memset   element   turn   cin   

原文地址：https://www.cnblogs.com/asunayi/p/13131125.html