给定一段伪代码,求程序结束后res的值
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
如果cur<0时就会break,由于每次操作只会加一或者减一,每次break时cur的初始值都会加一,所以每次break是具有单调性的。我们可以用一个变量来记录cur的最小值,因为只有小于这个最小值时才会break,所以当每次小于最小值时计算ans的值。
#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
void solve(){
string s;cin>>s;
int n=s.size();
int cur=0,worst=0;
ll ans=0;
rep(i,0,n){
if(s[i]=='+') cur++;
else cur--;
if(cur<worst){
ans+=i+1;
worst=cur;
}
}
ans+=n;
cout<<ans<<endl;
}
int main(){
IOS;
int t;cin>>t;
while(t--){
solve();
}
return 0;
}