CodeForces 940D

题目链接

https://vjudge.net/problem/CodeForces-940D

题意

给数组a和b，还有l和r的限制条件，输出符合条件的任意一组l和r。

思路

根据条件模拟。

AC代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<list>
#include<cstdlib>
#include<ctime>
using namespace std;
typedef long long ll;
const double PI=atan(1.0)*4;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll MOD = 1000000007;
const int maxn = 100010;

int a[maxn], b[maxn];
char s[maxn];

int main()
{
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    int n, l, r, x;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    scanf("%s", s);
    for(int i = 0; i < n; i++)
        b[i] = s[i] - '0';
    l = -1000000000;
    r = 1000000000;
    for(int i = 4; i < n; i++)
    {
        if(b[i-1] && b[i-2] && b[i-3] && b[i-4] && !b[i])
        {
            x = min(a[i], a[i - 1]);
            x = min(x, a[i - 2]);
            x = min(x, a[i - 3]);
            x = min(x, a[i - 4]);
            x--;
            r = min(x, r);
        }
        else if(!b[i-1] && !b[i-2] && !b[i-3] && !b[i-4] && b[i])
        {
            x = max(a[i], a[i - 1]);
            x = max(x, a[i - 2]);
            x = max(x, a[i - 3]);
            x = max(x, a[i - 4]);
            x++;
            l = max(x, l);
        }
    }
    printf("%d %d\n", l, r);
    return 0;
}