D. Program

You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
increase x by 1;
decrease x by 1.
You are given m queries of the following format: query l r — how many
distinct values is x assigned to if all the instructions between the

l-th one and the r-th one inclusive are ignored and the rest are
executed without changing the order?

Input The first line contains a single integer t (1≤t≤1000) — the
number of testcases. Then the description of t testcases follows. The
first line of each testcase contains two integers n and m
(1≤n,m≤2⋅105) — the number of instructions in the program and the

number of queries. The second line of each testcase contains a program
— a string of n characters: each character is either ‘+’ or ‘-’ —
increment and decrement instruction, respectively. Each of the next m
lines contains two integers l and r (1≤l≤r≤n) — the description of the
query. The sum of n over all testcases doesn’t exceed 2⋅105. The sum

of m over all testcases doesn’t exceed 2⋅105.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.

題意：一個有‘+’和‘-’組成的字串 初始x的值為0，然後開始執行字串，遇到‘+’ x的值+1，遇到‘-’ x的值-1 ，每次給你一個區間[L,R]，忽略區間內的字元，執行其他部分，問每次執行後，會出現多少個不同的數字

思路:忽略[L,R]內的字串，保留了兩部分[1,L] 和 [R,N] 因為x的值每次都是+1或者-1，所以x的值一定都是連續出現的，區間內的min–max之間的所有值都會全部出現，講所有字串執行得到num陣列然後用RMQ維護一下，ST表查詢兩邊區間的最大最小值，得到兩個答案區間範圍，很簡單可以找到右邊忽略區間[L,R]區間後的真實值 然後將兩個區間取並集，區間長度就是答案。

注意處理邊界L=1和R=N;

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=2e5+7;
int num[maxn],n,m,dp1[maxn][20],dp2[maxn][20];
void RMQ()
{
    for(int i=1;i<=n;i++)
        dp1[i][0]=dp2[i][0]=num[i];
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i+(1<<j)-1<=n;i++)
        {
            dp1[i][j]=max(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
            dp2[i][j]=min(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
        }
    }
}
int ans1=0,ans2=0;
void ST(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1)
    k++;
    ans1=max(dp1[l][k],dp1[r-(1<<k)+1][k]);
    ans2=min(dp2[l][k],dp2[r-(1<<k)+1][k]);
}
char s[maxn];
int main()
{
	int T,q;
	int l,r;
	cin>>T;
	while(T--)
	{
		cin>>n>>q;
		cin>>s+1;
		int cnt=0;
		for(int i=1;i<=n;i++)
		{
			if(s[i]=='+')
				num[i]=(++cnt);
			else num[i]=(--cnt);
		}
		RMQ();
		while(q--)
		{
			cin>>l>>r;
			int w1,w2,w3,w4;//記錄最大最小值 
			ans1=0,ans2=0;
			if(l-1>=1)
				ST(1,l-1);
			w1=max(ans1,0);
			w2=min(ans2,0);
			if(r+1<=n)
			{
				ST(r+1,n);
				int t;
				if(s[r+1]=='+')
				{
					int t=num[r+1]-1;
	
				}
				else{
					int t=num[r+1]+1;
				}
				t=num[r]-num[l-1];
				ans1=ans1-t;
				ans2=ans2-t;
			}
				
			else ans1=0,ans2=0;
			//w1-w2  ans1-ans2
			int ppt;
			if(ans2>w1||w2>ans1)
				ppt=(w1-w2+1)+(ans1-ans2+1);
			else {
				ppt=max(w1,ans1)-min(w2,ans2)+1;
			}
			cout<<ppt<<endl;
		}
		
	}
}

還剩幾分鐘寫出來這道題，看到這執行時間感覺會被HACK掉，結果還是活下來了，完結撒花，早日上綠~~