模拟AtCoDeer and Rock-Paper

Posted 2020-11-09 听风不成泣

题目描述

AtCoDeer the deer and his friend TopCoDeer is playing a game. The game consists of N turns. In each turn, each player plays one of the two gestures, Rock and Paper, as in Rock-paper-scissors, under the following condition:
(※) After each turn, (the number of times the player has played Paper)≤(the number of times the player has played Rock).
Each player‘s score is calculated by (the number of turns where the player wins) − (the number of turns where the player loses), where the outcome of each turn is determined by the rules of Rock-paper-scissors.
(For those who are not familiar with Rock-paper-scissors: If one player plays Rock and the other plays Paper, the latter player will win and the former player will lose. If both players play the same gesture, the round is a tie and neither player will win nor lose.)
With his supernatural power, AtCoDeer was able to foresee the gesture that TopCoDeer will play in each of the N turns, before the game starts. Plan AtCoDeer‘s gesture in each turn to maximize AtCoDeer‘s score.
The gesture that TopCoDeer will play in each turn is given by a string s. If the i-th (1≤i≤N) character in s is g, TopCoDeer will play Rock in the i-th turn. Similarly, if the i-th (1≤i≤N) character of s in p, TopCoDeer will play Paper in the i-th turn.

Constraints
1≤N≤10⁵
N=|s|
Each character in s is g or p.
The gestures represented by s satisfy the condition (※).

输入

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

输出

Print the AtCoDeer‘s maximum possible score.

样例输入

gpg

样例输出

0

提示

Playing the same gesture as the opponent in each turn results in the score of 0, which is the maximum possible score.

来源

ABC046&ARC062

 1 #include <iostream>
 2 #include <bits/stdc++.h>
 3 using namespace std;
 4 string a;
 5 int main()
 6 {
 7     cin>>a;
 8     int ans=0;
 9     int p=0;
10     for(int i=0;i<a.size();i++)
11     {
12         if(a[i]==‘g‘)
13         {
14             if(p>0)
15             {
16                 ans++;
17                 p--;
18             }
19             else
20             {
21                 p++;
22             }
23         }
24         else
25         {
26             if(p>0)
27                 p--;
28             else
29             {
30                 p++;
31                 ans--;
32             }
33 
34         }
35     }
36     cout << ans << endl;
37     return 0;
38 }

View Code