C. Doremy’s IQ
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Doremy is asked to test n contests. Contest i can only be tested on day i. The difficulty of contest i is ai. Initially, Doremy’s IQ is q. On day i Doremy will choose whether to test contest i or not. She can only test a contest if her current IQ is strictly greater than 0.
If Doremy chooses to test contest i on day i, the following happens:
if ai>q, Doremy will feel she is not wise enough, so q decreases by 1;
otherwise, nothing changes.
If she chooses not to test a contest, nothing changes.
Doremy wants to test as many contests as possible. Please give Doremy a solution.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of the test cases follows.
The first line contains two integers n and q (1≤n≤105, 1≤q≤109) — the number of contests and Doremy’s IQ in the beginning.
The second line contains n integers a1,a2,⋯,an (1≤ai≤109) — the difficulty of each contest.
It is guaranteed that the sum of n over all test cases does not exceed 105.
Output
For each test case, you need to output a binary string s, where si=1 if Doremy should choose to test contest i, and si=0 otherwise. The number of ones in the string should be maximum possible, and she should never test a contest when her IQ is zero or less.
If there are multiple solutions, you may output any.
Example
inputCopy
5
1 1
1
2 1
1 2
3 1
1 2 1
4 2
1 4 3 1
5 2
5 1 2 4 3
outputCopy
1
11
110
1110
01111
Note
In the first test case, Doremy tests the only contest. Her IQ doesn’t decrease.
In the second test case, Doremy tests both contests. Her IQ decreases by 1 after testing contest 2.
In the third test case, Doremy tests contest 1 and 2. Her IQ decreases to 0 after testing contest 2, so she can’t test contest 3.
原题链接：C. Doremy’s IQ
思路：这题顺着跑比较难想，因为IQ是递减的（越写题越傻），所以思路改为从后往前跑，每次解出来的都是最优的，逆着推IQ值。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<map>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define int long long
#define debug(x) cerr << #x << ": " << (x) << endl
#define lowbit(i) (-i)&i
#define mem(i) memset(i,0,sizeof i)
using namespace std;
const int N = 5e5;
int n,m,t;
signed main(){
IOS
while(cin >> t){
while(t--){
cin >> n >> m;
vector<int> ve(n),a(n,0);
for(int i = 0 ; i < n ; i ++)cin >> ve[i];
int IQ = 0;
for(int i = n - 1 ; i >= 0 ; i --){
if(IQ >= ve[i]){
//智商足够时，赢得这场比赛
a[i] = 1;
}else if(IQ < m){
//当智商不够时，消费1智商得到这场比赛
a[i] = 1;
IQ++;
}
}
for(int i = 0 ; i < n ; i ++)cout << a[i];
cout << endl;
}
}
return 0;
}
