2019 cattle off more school first A-Equivalent Prefixes

Original title Address

Portal

Problem-solving ideas

First monotonous stack get two sequences is smaller than the first number of each of the left own index number, stored in a [], b []. Then the order of 1 ~ n cycles, comparing a [i] and b [i] are equal to, if not equal loop is exited, at this time is equal to a final answer.
When the former is assumed that 1 ~ n-1 conditions have been satisfied, then determining the feasibility of 1 ~ n, l ~ n is determined whether both satisfied, if a [n] <[n] b, then when l = b [n] , RBQ sequence is b [n], a sequence of n, obviously does not hold 2, a [n]> b [ n] the same way. When a [n] is equal to b [n], l> a [n] are RBQ case n, l <= n when it has been possible because of the smallest number n, there is no effect on the front, clearly established a.

code show as below

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

inline int read(){
    int res = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -res : res;
}

const int N = 100005;

struct T{
    int val, i;
    T(int val, int i): val(val), i(i){}
};
stack<T> sta1, sta2;
int a[N], b[N];


int main()
{
    int n;
    while(scanf("%d", &n) != EOF){
        for(int i = 1; i <= n; i ++){
            int x = read();
            while(!sta1.empty() && sta1.top().val > x)
                sta1.pop();
            if(!sta1.empty())
                a[i] = sta1.top().i;
            else
                a[i] = 0;
            sta1.push(T(x, i));
        }
        for(int i = 1; i <= n; i ++){
            int x = read();
            while(!sta2.empty() && sta2.top().val > x)
                sta2.pop();
            if(!sta2.empty())
                b[i] = sta2.top().i;
            else
                b[i] = 0;
            sta2.push(T(x, i));
        }
        while(!sta1.empty())
            sta1.pop();
        while(!sta2.empty())
            sta2.pop();
        int ans = 0;
        for(int i = 1; i <= n; i ++){
            if(a[i] == b[i])
                ans = i;
            else
                break;
        }
        printf("%d\n", ans);
    }
    return 0;
}