6486: An Ordinary Game
时间限制: 1 Sec 内存限制: 128 MB提交: 150 解决: 75
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题目描述
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
3≤|s|≤105
s consists of lowercase English letters.
No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
3≤|s|≤105
s consists of lowercase English letters.
No two neighboring characters in s are equal.
输入
The input is given from Standard Input in the following format:
s
s
输出
If Takahashi will win, print First. If Aoki will win, print Second.
样例输入
aba
样例输出
Second
提示
Takahashi, who goes first, cannot perform the operation, since removal of the b, which is the only character not at either ends of s, would result in s becoming aa, with two as neighboring.
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int main()
{
char s[100005], ss[100005];
scanf("%s", s);
int sum=0, ans=0, len=strlen(s);
while(1)
{
for(int i=1;i<len-1;i++)
{
if(s[i]!='\0')
{
int x=i-1;
while(s[x]=='\0') x--;
if(s[x]!=s[i+1])
{
s[i]='\0';
sum++;
}
}
}
int cnt=0;
for(int i=0;i<len;i++)
if(s[i]!='\0')
ss[cnt++]=s[i];
for(int i=0;i<len;i++)
s[i]=ss[i];
len=cnt;
if(sum==ans)
break;
ans=sum;
}
if(sum%2)
printf("First\n");
else
printf("Second\n");
return 0;
}
/**************************************************************
Problem: 6486
User: ldu_reserver201701
Language: C++
Result: 正确
Time:4 ms
Memory:1168 kb
****************************************************************/