Gauss FibonacciTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2584 Accepted Submission(s)
Gauss Fibonacci
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2584 Accepted Submission(s): 1078
Problem Description
Without expecting,Angel replied quickly.She says: “I’v heard that you’r a very clever boy. So if you wanna me be your GF,you should solve the problem called GF~. ”
How good an opportunity that Gardon can not give up! The “Problem GF” told by Angel is actually “Gauss Fibonacci”.
As we know,Gauss is the famous mathematician who worked out the sum from 1 to 100 very quickly,and Fibonacci is the crazy man who invented some numbers.
Arithmetic progression:
g(i)=k*i+b;
We assume k and b are both non-nagetive integers.
Fibonacci Numbers:
f(0)=0
f(1)=1
f(n)=f(n⑴)+f(n⑵) (n>=2)
The Gauss Fibonacci problem is described as follows:
Given k,b,n,calculate the sum of every f(g(i)) for 0<=i
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define LL __int64
struct node{
LL num[3][3];
};
node e,a;
LL n,m,k,b;
node mul(node aa,node bb){
node c;
for(int i = 1; i < 3; ++i){
for(int j = 1; j < 3; ++j){
c.num[i][j] = 0;
for(int k = 1; k < 3; ++k){
c.num[i][j] = (c.num[i][j]+aa.num[i][k]*bb.num[k][j])%m;
}
}
}
return c;
}
node fa(node a,LL n){
node b = e;
while(n){
if(n&1) b = mul(a,b);
n >>= 1;
a = mul(a,a);
}
return b;
}
node add(node aa,node bb){
node c;
for(int i = 1; i < 3; ++i){
for(int j = 1; j < 3; ++j){
c.num[i][j] = (aa.num[i][j]+bb.num[i][j])%m;
}
}
return c;
}
node dg(node p,LL k){
if(k == 1) return p;
else if(k&1) return add(dg(p,k-1),fa(p,k));
else return mul(dg(p,k>>1),add(fa(p,e));
}
int main(){
e.num[1][1] = e.num[2][2] = 1;
e.num[1][2] = e.num[2][1] = 0;
a.num[1][1] = a.num[1][2] = a.num[2][1] = 1; a.num[2][2] = 0;
while(cin >> k >> b >> n >> m){
node ak = fa(a,k);
node ab = fa(a,b);
node ans = dg(ak,n-1);
ans = add(e,ans);
ans = mul(ab,ans);
cout << ans.num[1][2]<< endl;
}
return 0;
}
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