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

Pluses and Minuses

题目大意：（文末有原题）

给出一个有'+' '-'组成的字符串，读伪代码写程序；

res = 0
for init = 0 to inf（0到正无穷，其实就是判断条件为1）
    cur = init 
    ok = true 
    for i = 1 to |s|
        res = res + 1 （每进入一次循环，res++）
        if s[i] == '+'
            cur = cur + 1
        else
            cur = cur - 1
        if cur < 0 （如果cur<0会跳出）
            ok = false
            break
    if ok   （如果cur到最后>=0，就出循环）
        break

要求输出跳出循环后res的值，即循环多少次，能够使cur通过一轮循环后>=0（cur每次从cur+1开始）;

思路：（按照伪代码暴力写下来程序会TLE）

其实就是输出能够进行多少次循环，因为如果此步>0，就会进行下一步，所以只有当遇到<0，才会再次从头开始循环，所以我们只需记录每一次小于0，需要进行几步，并求出最多能走几步而不跳出循环，此时，再循环一次，就会直接出循环；所以，我们只需将这些步数加起来再加上字符串长度（最后一次直接从头走到尾）得到的就是循环次数；

代码：

#include 
#include 
#include 

原题：

题目：

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.

输入：

The first line contains one integer t (1≤t≤1000) — the number of test cases.

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

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

输出：

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

样例：

Input：

3
--+-
---
++--+-

Output：

7
9
6