标题：rating 赛 补提

B. Beautiful Numbers
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a permutation p=[p1,p2,…,pn] of integers from 1 to n. Let’s call the number m (1≤m≤n) beautiful, if there exists two indices l,r (1≤l≤r≤n), such that the numbers [pl,pl+1,…,pr] is a permutation of numbers 1,2,…,m.

For example, let p=[4,5,1,3,2,6]. In this case, the numbers 1,3,5,6 are beautiful and 2,4 are not. It is because:

if l=3 and r=3 we will have a permutation [1] for m=1;
if l=3 and r=5 we will have a permutation [1,3,2] for m=3;
if l=1 and r=5 we will have a permutation [4,5,1,3,2] for m=5;
if l=1 and r=6 we will have a permutation [4,5,1,3,2,6] for m=6;
it is impossible to take some l and r, such that [pl,pl+1,…,pr] is a permutation of numbers 1,2,…,m for m=2 and for m=4.
You are given a permutation p=[p1,p2,…,pn]. For all m (1≤m≤n) determine if it is a beautiful number or not.

Input
The first line contains the only integer t (1≤t≤1000) — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number n (1≤n≤2⋅105) — the length of the given permutation p. The next line contains n integers p1,p2,…,pn (1≤pi≤n, all pi are different) — the given permutation p.

It is guaranteed, that the sum of n from all test cases in the input doesn’t exceed 2⋅105.

Output
Print t lines — the answers to test cases in the order they are given in the input.

The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
struct ty {
	int val;
	int pos;
};
struct ty p[200005];
int t;
int n;
int cmp(const void* a, const void* b) {
	return ((struct ty*)a)->val - ((struct ty*)b)->val;
}
int max_(int x, int y) {
	if (x < y) return y;
	return x;
}
int min_(int x, int y) {
	if (x > y) return y;
	return x;
}
int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) {
			scanf("%d", &p[i].val);
			p[i].pos = i;
		}
		qsort(&p[1], n, sizeof(p[1]), cmp);
		printf("1");
		int max = p[1].pos;
		int min = p[1].pos;
		for (int i = 2; i <= n; i++) {
			max =max_ (max, p[i].pos);
			min =min_ (min, p[i].pos);
			if (max - min + 1 == i) printf("1");
			else printf("0");
		}
		printf("n");
	}
	return 0;
}

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