题意概述:给你n个数p[i],代表第i个右括号之前的左括号个数,让你输出一个w数组。w[i]代表与其相匹配的左括号的区间内的右括号的个数
思想:我们可以先根据p数组模拟出这个字符串,然后在由字符串反推出w数组
代码实现
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<bitset>
#include<math.h>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define PI acos(-1)
#define close ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define open #ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);freopen("out.txt","w",stdout); #endif
using namespace std;
typedef long long ll;
const double pai=3.141592653589793238462643383279;
const int MAX_N = 1000000+50;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-10;
const ll INF_ll = 0x7fffffffffffffff;
ll mod = 1e9+7;
/*
不要放弃
Don't give up
心中有梦(ICPC)
*/
inline int read()
{
int x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9') {
if(ch=='-') f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9') {
x=10*x+ch-'0';
ch=getchar();
}
return x*f;
}
inline void Out(int a)
{
if(a>9)
Out(a/10);
putchar(a%10+'0');
}
multiset<int> s;
int p[MAX_N],vis[MAX_N],w[MAX_N];
int main()
{
string s;
int t,n;
cin>>t;
while(t--){
cin>>n;
s="";
for(int i = 1; i <= n; i++) cin>>p[i];
p[0] = 0;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= (p[i]-p[i-1]); j++)
s+="(";
s+=")";
}
memset(vis,0,sizeof(vis));
int k = 1;
for(int i = 0; i < 2*n; i++){
int ans = 1;
if(s[i] == ')'){
for(int j = i-1; j >= 0; j--){
if(s[j] == ')'){
ans++;
}
if(s[j] == '(' && !vis[j]){
vis[j] = 1;
break;
}
}
w[k++] = ans;
}
}
for(int i = 1; i <= n; i++){
cout<<w[i]<<" ";
}
cout<<endl;
}
return 0;
}
/*
********
************
####....#.
#..###.....##....
###.......###### ### ###
........... #...# #...#
##*####### #.#.# #.#.#
####*******###### #.#.# #.#.#
...#***.****.*###.... #...# #...#
....**********##..... ### ###
....**** *****....
#### ####
###### ######
##############################################################
#...#......#.##...#......#.##...#......#.##------------------#
###########################################------------------#
#..#....#....##..#....#....##..#....#....#####################
########################################## #----------#
#.....#......##.....#......##.....#......# #----------#
########################################## #----------#
#.#..#....#..##.#..#....#..##.#..#....#..# #----------#
########################################## ############
*/