CodeForces 1579F :Array Stabilization (AND version) 思维

传送门

题意

给定一个数组 n n n，下标从 0 0 0开始，只包含 0 0 0和 1 1 1.给定 d d d。现在需要你确定经过多少次操作后数组元素全为0。
操作为将数组循环移位 d d d，再和原数组相与得到新数组。

分析

首先0和任何数操作完都是0，所以只要考虑将0移动1的位置即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e6 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int n,a[N],d;

int solve(){
    int res = 0;
    queue<PII> q;
    for(int i = 0;i < n;i++) if(!a[i]) q.push({0,i});  
    while(q.size()){
        auto t = q.front();
        q.pop();
        int x = (t.se + d) % n;
        if(a[x]){
            a[x] = 0;
            q.push({t.fi + 1,x});
            res = max(res,t.fi + 1);
        }
    }
    for(int i = 0;i < n;i++) if(a[i]) res = -1;
    return res; 
}

int main() {
    int T;
    read(T);
    while(T--){
        read(n),read(d);
        for(int i = 0;i < n;i++) read(a[i]);
        di(solve());
    }
    return 0;
}

标签：1579F,Stabilization,int,res,CodeForces,read,while,数组,const
来源： https://blog.csdn.net/tlyzxc/article/details/120616472