AtCoder Grand Contest 015（B - Evilator）

Problem Statement

Skenu constructed a building that has N floors. The building has an elevator that stops at every floor.

There are buttons to control the elevator, but Skenu thoughtlessly installed only one button on each floor - up or down. This means that, from each floor, one can only go in one direction. If S_i is U, only "up" button is installed on the i-th floor and one can only go up; if S_i is D, only "down" button is installed on the i-th floor and one can only go down.

The residents have no choice but to go to their destination floors via other floors if necessary. Find the sum of the following numbers over all ordered pairs of two floors (i,j): the minimum number of times one needs to take the elevator to get to the j-th floor from the i-th floor.

Constraints

• 2≤|S|≤10⁵
• S_i is either U or D.
• S₁ is U.
• S_|S| is D.

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Input

The input is given from Standard Input in the following format:

S1S2…S|S|

Output

Print the sum of the following numbers over all ordered pairs of two floors (i,j): the minimum number of times one needs to take the elevator to get to the j-th floor from the i-th floor.

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Sample Input 1

Copy

UUD

Sample Output 1

Copy

7

From the 1-st floor, one can get to the 2-nd floor by taking the elevator once.

From the 1-st floor, one can get to the 3-rd floor by taking the elevator once.

From the 2-nd floor, one can get to the 1-st floor by taking the elevator twice.

From the 2-nd floor, one can get to the 3-rd floor by taking the elevator once.

From the 3-rd floor, one can get to the 1-st floor by taking the elevator once.

From the 3-rd floor, one can get to the 2-nd floor by taking the elevator once.

The sum of these numbers of times, 7, should be printed.

 

英语水平有待提高，这题是看例子猜题意的。５５５５５５

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
char a[N];
int b[N];
int main(){
    cin >> a;
    int flag = 0, flag1 = 0;
    ll ans = 0, len = strlen(a);
    for(int i = len-1; i >= 0; i--){
        if(a[i] == 'D'){
            flag = 1;
        }
        if(flag) b[i] = 1;
        else b[i] = 0;
    }
    for(int i = 0; i < len; i ++){
        if(a[i] == 'U'){
            flag1 = 1;
            ans += (len-i-1);
            if(b[i]){
                ans += 2*i;
            }
        }else if(a[i] == 'D'){
            ans += i;
            if(flag1){
                ans += 2*(len-i-1);
            }
        }
    }
    cout << ans << endl;
    return 0;
}