codeforces div1 B / codeforces div2 D
2 seconds
256 megabytes
standard input
standard output
Fibonacci numbers are the sequence of integers: f0 = 0, f1 = 1, f2 = 1, f3 = 2, f4 = 3, f5 = 5, ..., fn = fn - 2 + fn - 1.
So every next number is the sum of the previous two.
Bajtek has developed a nice way to compute Fibonacci numbers on a blackboard. First, he writes a 0. Then, below it, he writes a 1. Then he performs the following two operations:
- operation "T": replace the top number with the sum of both numbers;
- operation "B": replace the bottom number with the sum of both numbers.
If he performs n operations, starting with "T" and then
choosing operations alternately (so that the sequence of operations looks like "TBTBTBTB..."),
the last number written will be equal to fn + 1.
Unfortunately, Bajtek sometimes makes mistakes and repeats an operation two or more times in a row. For example, if Bajtek wanted to compute f7,
then he would want to do n = 6 operations: "TBTBTB".
If he instead performs the sequence of operations "TTTBBT", then he will have made 3 mistakes, and he will incorrectly compute that the seventh Fibonacci number
is 10. The number of mistakes in the sequence of operations is the number of neighbouring equal operations («TT» or «BB»).
You are given the number n of operations that Bajtek has made in an attempt to compute fn + 1 and
the number r that is the result of his computations (that is last written number). Find the minimum possible number of mistakes that Bajtek must have made
and any possible sequence of n operations resulting inr with
that number of mistakes.
Assume that Bajtek always correctly starts with operation "T".
The first line contains the integers n and r (1 ≤ n, r ≤ 106).
The first line of the output should contain one number — the minimum possible number of mistakes made by Bajtek. The second line should contain n characters,
starting with "T", describing one possible sequence of operations with that number of mistakes. Each character must be either "T"
or "B".
If the required sequence doesn't exist, output "IMPOSSIBLE" (without quotes).
6 10
2 TBBTTB
4 5
0 TBTB
2 1
IMPOSSIBLE
比赛的时候没想清楚，后来发现其实很简单的。
最后要到达数字r，那么另一个数一定比r小，且另一个数确定之后，那么转移过程也就确定了，所以就可以枚举另外这个数，然后模拟转移过程判断符不符合条件，如果符合条件的话，再保存路径输出结果就行了。
转移的过程如果遇上要一直覆盖某一个数，那就直接用除法，而不一步步模拟，这样就会很快了。
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define INF 100000000
int n,s,num;
char ans[1000005];
void Check(int x,int y)
{
int p,t=0;
int k=0;
int ret=x;
while(1)
{
if (x<y)
{
p=x;
x=y;
y=p;
}
if (x==y || y==0) break;
k++;
t+=x/y;
x=x%y;
}
if (x!=1 || y!=0) return;
if (t==n && s<k)
{
s=k;
num=ret;
}
}
int main()
{
int i,j,t,r;
scanf("%d%d",&n,&r);
if (n==1 && r==1)
{
printf("0\nT\n");
return 0;
}
s=-INF;
for (i=1;i<r;i++)
{
Check(i,r);
}
if (s==-INF)
{
printf("IMPOSSIBLE\n");
return 0;
}
printf("%d\n",n-s-1);
int x=num,y=r,up=0;
while(1)
{
if (x==1 && y==1) break;
if (x<y)
{
ans[up++]='T';
y-=x;
}
else
{
ans[up++]='B';
x-=y;
}
}
printf("T");
for (i=up-1;i>=0;i--)
{
if (ans[up-1]=='B') printf("%c",ans[i]);
else printf("%c",ans[i]=='T'?'B':'T');
}
return 0;
}