Single Point Test time: 1.0 seconds
Memory Limit: 512 MB
This is a C / C ++ code for calculating the recurrence sequence an.
const int MOD = (int)1e9 + 7; int a(int n) { if(n < 3) return 1; return ((a(n - 1) << 1) % MOD + (a(n - 2) >> 1)) % MOD; }
Please an output value of n items according to the number of columns in the code.
Entry
A first line integer T.
T lines after each line an integer representing n.
0 <T≤105, 0 <n≤106.
Export
Please each line of an output value of one.
Sample
input
3 1 2 3
output
1 1 2
直接递归一定会超时所以可直接转化为数学式子来处理
#include<bits/stdc++.h>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MOD = (int)1e9 + 7;
ll b[1000005];
/*ll a(ll n) {
if(n < 3) return 1;
return
((a(n - 1)%MOD << 1) % MOD + (a(n - 2)%MOD >> 1)) % MOD;
}*/
/*ll a(ll n) {
if(n < 3) return b[n];
int res = ((a(n - 1)%MOD << 1) % MOD + (a(n - 2)%MOD >> 1)) % MOD;
return b[n] = res;
}*/
int main()
{
int t;
scanf("%d",&t);
b[1] = b[2] = 1;
for (int i=3;i<=1000000;i++)
{
b[i] = (b[i-1] * 2 % MOD +b [i-2] / 2 % MOD)%MOD;
}
while (t--)
{
ll n;
scanf("%lld",&n);
printf("%lld\n",b[n]);
}
return 0;
}