2018 Beijing ICPC D. Frog and Portal Fibonacci structure

Topic link: http: //hihocoder.com/contest/icpcbeijing2018/problem/4

Meaning of the questions: frog can jump from p p + 1, p + 2, now the frog jumps from 0 200, you can install the portal ai-aj a frog will jump to ai aj, may form a loop, each Portals ai different,

After seeking security portal, frogs jump program number is exactly 200 m of placement method

answer:

I did not do it. . We see the solution to a problem, I feel an ingenious

flat:

Any positive integer that can be broken down into a number of different Fibonacci number Pact and

f [48] << 1 has greater than 32, and can be subtracted from the Fibonacci descending, even upside down when scored, such as from 199-200 = 1,198-200 = 2,197-200 = 3 ......, and even 1 197,3 198,5 even 197 connected end to end their own node even 6-6

A front spacer connected to a rear, such as front of each embodiment 1, is not connected to the back of the impact on the back.

Even as the first 1-195, 3-198 and even a, that is equivalent to a 1 in 198 plus program, 199 program also added 1, 200 plus 2 program, but also a fresh start Fibonacci superimposed He scored on a front.

Figure

planb: If we at x, programs for m, m is an even number: Connection x + 1 ~ x + 3, x + 2 ~ x + 3, x + 3 to this embodiment there are two kinds, the problem becomes to x + 3 200, programs for m / 2,

m is an odd number, even x + 1 ~ 200, becomes a problem to x + 2 m 1-200 Proposed program number of the program

Sentenced to 0,1 special circumstances

AC Code:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int const maxn=210;
 4 typedef long long ll;
 5 long long f[maxn],m,tot,ans[maxn];
 6 void getfi(){
 7     f[200]=1;f[199]=1;
 8     for(int i=198;i>=150;i--)f[i]=f[i+1]+f[i+2];
 9 }
10 void work(){
11     if(m==0){
12         printf("2\n1 1\n2 1\n");return ;
13     }
14     if(m==1){
15         printf("2\n1 199\n2 2\n");return ;
16     }
17     if(m==2){printf("2\n1 2\n2 199\n");return ;}
18     for(ll i=150;i<=199;i++){
19         if(m==0)break;
20         if(m-f[i]>=0){
21             ans[++tot]=i;
22             m-=f[i];
23         }
24     }
25     printf("%lld\n",tot+1);
26     for(int i=1;i<=tot;i++)printf("%d %lld\n",i*2-1,ans[i]);
27     printf("%lld %lld\n",tot*2,tot*2);
28 }
29 int main(){
30     getfi();
31     while(scanf("%lld",&m)!=EOF){
32         tot=0;
33         work();
34     }
35     return 0;
36 }