HDU5950 Recursive sequence 非线性递推式 矩阵快速幂
题目描述：给出一个数列的第一项和第二项，计算第n项。
递推式是 f(n)=f(n-1)+2*f(n-2)+n^4.
由于n很大，所以肯定是矩阵快速幂的题目，但是矩阵快速幂只能解决线性的问题，n^4在这个式子中是非线性的，后一项和前一项没有什么直接关系，所以模拟赛的时候想破头也不会做。
这里要做一个转换，把n^4变成一个线性的，也就是和（n-1）^4有关系的东西，而这个办法就是：
n^4=(n-1+1)^4=(n-1)^4+4*(n-1)^3+6*(n-1)^2+4*(n-1)^1+(n-1)^0;
这个转换就建立了某一项和前一项的关系，矩阵的F数组就是 f[7]={ b , a , 81 , 27 , 9 , 3 , 1 };，整体的矩阵也很好构造，代码里有。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<sstream>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<bitset>
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
ll mod=2147493647;
inline int rd(void) {
int x=0; int f=1;char s=getchar();
while(s<'0'||s>'9') { if(s=='-')f=-1; s=getchar();}
while(s>='0'&&s<='9') { x=x*10+s-'0';s=getchar();}
x*=f;return x;
}
ll n,a,b;
void mul(ll f[7],ll a[7][7]){
ll c[7];
CLR(c,0);
for(int j=0;j<7;j++){
for(int k=0;k<7;k++){
c[j]=(c[j]+f[k]*a[k][j]%mod)%mod;
}
}
memcpy(f,c,sizeof(c));
}
void mulself(ll a[7][7]){
ll c[7][7];
CLR(c,0);
for(int i=0;i<7;i++){
for(int j=0;j<7;j++){
for(int k=0;k<7;k++){
c[i][j]=(c[i][j]+a[i][k]*a[k][j]%mod)%mod;
}
}
}
memcpy(a,c,sizeof(c));
}
int main(){
int T;
cin>>T;
while(T--){
scanf("%lld%lld%lld",&n,&a,&b);
if(n==1){
printf("%lld\n",a);
}else if(n==2){
printf("%lld\n",b);
}else{
ll f[7]={b,a,81,27,9,3,1};
ll a[7][7]=
{{1,1,0,0,0,0,0},
{2,0,0,0,0,0,0},
{1,0,1,0,0,0,0},
{0,0,4,1,0,0,0},
{0,0,6,3,1,0,0},
{0,0,4,3,2,1,0},
{0,0,1,1,1,1,1}};
n-=2;
for(;n;n>>=1){
if(n&1)mul(f,a);
mulself(a);
}
printf("%lld\n",f[0]);
}
}
}Recursive sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3374 Accepted Submission(s): 1485
Problem Description
Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right.
Input
The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.
Output
For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.
Sample Input
2 3 1 2 4 1 10
Sample Output
85 369
Hint
In the first case, the third number is 85 = 2*1十2十3^4. In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.
