CodeForces - 626B Cards (全排列&模拟)
| CodeForces - 626B
Cards
Description Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
She repeats this process until there is only one card left. What are the possible colors for the final card? Input The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards. The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively. Output Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order. Sample Input
Input
2 RB
Output
G
Input
3 GRG
Output
BR
Input
5 BBBBB
Output
B Hint In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card. In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card. In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. //题意： 给你n个气球，这n个气球有三种颜色，蓝色（B），绿色（G），红色（R），现在玩一个游戏，每次从n个气球中拿出来两个，如果这两个气球颜色不同，那么这两个气球可以合并成一个第三种颜色的气球，如果这两个气球颜色相同，那么这两个气球可以合并成一个同种颜色的气球，问这样一直合并，直到最后一个。问最后能合并出哪几种颜色的气球。 //思路： 先将n个颜色不同的气球的颜色转换成数字，并将其存在一个数组a中，对a数组进行全排列，并对每一种情况进行合并，直到最后。（当合并出来的气球颜色已经等于3时，直接结束，节省时间，因为总共就3 种颜色）。
//HAIT： 首先得说一下输出格式，有的输入会对应多种输出，但要输出按字母升序排列（WA了3次）。。。 另外用我的思路写有一个bug，那就是当输入为（4 BGBG）时，我的输出是BG，但答案应该是BGR，在这块WA了8次。。。但加了一个特殊判断就过了^_^（在下面代码中会标注）。。。 #include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 100010
#define M 1000000007
using namespace std;
char s[210];
int a[210];
int b[210];
int vis[5];
int c[5];
int d[5];
int chang(char c)
{
if(c=='B') return 1;
if(c=='G') return 2;
if(c=='R') return 3;
}
void ch(int x)
{
if(x==1) printf("B");
else if(x==2) printf("G");
else if(x==3) printf("R");
}
int main()
{
int n,i,j,k;
while(scanf("%d",&n)!=EOF)
{
memset(vis,0,sizeof(vis));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
scanf("%s",s);
int kk=0;
for(i=0;i<n;i++)
{
a[i]=chang(s[i]);
if(!vis[a[i]])
{
vis[a[i]]=1;
kk++;
}
d[a[i]]++;
}
if((kk==2)&&(d[1]==2&&d[2]==2)||(d[1]==2&&d[3]==2)||(d[2]==2&&d[3]==2))//加一个特殊判断
{
printf("BGR\n");
continue;
}
sort(a,a+n);
int k=0;
memset(vis,0,sizeof(vis));
do
{
for(i=0;i<n;i++)
b[i]=a[i];
// for(i=0;i<n;i++)
// printf("%d ",b[i]);
// printf("\n");
for(i=1;i<n;i++)
{
if(b[i]==b[i-1])
b[i]=b[i-1];
else
{
b[i]=(b[i]+b[i-1])%4;
if(!b[i])
b[i]=2;
}
}
// printf("%d\n",b[n-1]);
if(!vis[b[n-1]])
{
vis[b[n-1]]=1;
c[k++]=b[n-1];
}
if(k==3)
break;
}while(next_permutation(a,a+n));
sort(c,c+k);
for(i=0;i<k;i++)
ch(c[i]);
printf("\n");
}
return 0;
}
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