CF - 高精度 + 贪心

Posted 2020-10-24

Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:

• A customer came to Bob and asked to sell him a 2^x MB memory stick. If Bob had such a stick, he sold it and got 2^x berllars.
• Bob won some programming competition and got a 2^x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.

Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers‘ demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally.

Input

The first input line contains number n (1?≤?n?≤?5000) — amount of Bob‘s working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2^x MB memory stick (0?≤?x?≤?2000). It‘s guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2^x MB memory stick (0?≤?x?≤?2000).

Output

Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don‘t forget, please, that Bob can‘t keep more than one memory stick at a time.

Example

Input

7
win 10
win 5
win 3
sell 5
sell 3
win 10
sell 10

Output

1056

Input

3
win 5
sell 6
sell 4

Output

0

题目分析 ： 一个人卖内存条，sell 表示有人去买此内存条， win表示获取到此内存条，这个人最多只能拥有一个内存条，并且同时也保证买不同型号内存条的只会有一个顾客，问此人能获得的最大收益。
思路分析 ： 输出答案得是高精度，选购物品的时候贪心的去选， 因为 2^5 一定大于 2^1 + 2^2 + 2^3 + 2^4。
代码示例 ：

#define ll long long
const int maxn = 1e6+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int n;
struct node
{
    int state, x; // 1 是 win, 0 是 sell
    int pt; // 1 是存在
}pre[5005];
vector<int>f[2005];
int p[2005];
vector<int>ans, mid;

void init() {
    f[0].push_back(1);
    
    for(int i = 1; i <= 2000; i++){
        f[i].assign(f[i-1].size(), 0);
        for(int j = 0; j < f[i].size(); j++){
            f[i][j] = f[i-1][j] + f[i-1][j];
        }
        for(int j = f[i].size()-1; j >= 0; j--){
            if (f[i][j] > 9){
                f[i][j] -= 10;
                if (j != 0) f[i][j-1]++;
                else f[i].insert(f[i].begin(), 1);
            }
        }
    }    
}

void bigsum(int x){
    mid.clear();
    mid.assign(ans.size(), 0);
    
    for(int i = ans.size()-1, j = f[x].size()-1; i >= 0; ){
        if (j >= 0){
            mid[i] = ans[i] + f[x][j];
            i--, j--;
        }
        else {mid[i] = ans[i]; i--;}
    }
    
    for(int i = mid.size()-1; i >= 0; i--){
        if (mid[i] > 9){
            mid[i] -= 10;
            if (i != 0) mid[i-1]++;
            else mid.insert(mid.begin(), 1);
        }
    }
    ans = mid;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n;
    char s[10];
    int x;
    
    init();
    for(int i = 1; i <= n; i++) {
        scanf("%s%d", s, &x);
        if (s[0] == ‘w‘) pre[i].state = 1;
        else {
             pre[i].state = 0;
             p[x] = i;
        }
        pre[i].x = x; pre[i].pt = 1;        
    }
 
    int sign = 1;   
    for(int i = 2000; i >= 0; i--){
        if (!p[i]) continue;
        int pos = 999999;
        for(int j = p[i]-1; j >= 1; j--){
            if (pre[j].pt == 0) break;
            if (pre[j].state == 1 && pre[j].x == i){
                pos = j;
                if (sign) {ans = f[i]; sign = 0;}
                else bigsum(i);
                break;
                 
            }
        }
        for(int j = p[i]; j >= pos; j--){
            if (pre[j].state == 0) p[pre[j].x] = 0;
            else pre[j].pt = 0;
        } 
        //printf("-----  %d %d\n", i, pos);
    }
    if (sign) {printf("0\n"); return 0;} 
    for(int i = 0; i < ans.size(); i++){
        printf("%d", ans[i]);
    }
    printf("\n");
    //puts("");
    return 0;
}