Educational Codeforces Round 90 (Rated for Div. 2) C. Pluses and Minuses（题解）

You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:

res = 0
for init = 0 to inf
    cur = init
    ok = true
    for i = 1 to |s|
        res = res + 1
        if s[i] == '+'
            cur = cur + 1
        else
            cur = cur - 1
        if cur < 0
            ok = false
            break
    if ok
        break

Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.

You have to calculate the value of the res after the process ends.

Input
The first line contains one integer $t(1\le t\le 1000)$ — the number of test cases.

The only lines of each test case contains string s $(1\le |s|\le 10^6)$consisting only of characters + and -.

It’s guaranteed that sum of |s| over all test cases doesn’t exceed $10^6$.

Output
For each test case print one integer — the value of the res after the process ends.

题意：根据伪代码输出res，cur如果遇见–，则cur减1，cur为负数时，跳出循环，cur初始化cur加一。直到遍历s后，不为负数，输出res.
题解：- - + - : cur=0: s[0] cur=-1 break
cur=1: s[0] cur=0 s[1] cur=-1 break
cur=2: s[0] cur=1 s[1] cur=0 s[2] cur=1 s[3] cur =0
res =7
当cur<0 因此在j前面的都要重新循环 因此res+=j+1(j是从0开始),这时候 cur变为0.

// cf6.22.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//

#include 
#include
#define ll long long
using namespace std;
const int maxn = 1e6+10;
int main()
{
    int  t;
    cin >> t;
    while (t--) {
        string s;
        cin >> s;
        long long  res = 0;//int 会爆 注意
        int cur=0;
        for (int j = 0; j < s.size(); j++)
           {
              // res += 1;
               if (s[j] == '+') {
                   cur += 1;
               }
               else {
                   cur -= 1;
               }
               if (cur < 0) {
                   res += j + 1;//j是从0开始的所以加一
                   cur = 0;
               }
               
           }
        cout << res+s.size() << endl;//res是未到终点前的遍历，s.size()是最后一次到终点的遍历
    }
}