1052 - String Growth
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Time Limit: 2 second(s)Memory Limit: 32 MB
Zibon just started his courses in Computer science. After having some lectures on programming courses he fell in love with strings. He started to play with strings and experiments on them. One day he started a string of arbitrary (of course positive) length consisting of only {a, b}. He considered it as 1st string and generated subsequent strings from it by replacing all the b's with ab and all the a's with b. For example, if he ith string is abab, (i+1)th string will be b(ab)b(ab) = babbab. He found that the Nth string has length X and Mth string has length Y. He wondered what will be length of the Kth string. Can you help him?
Input
Input starts with an integer T (≤ 200), denoting the number of test cases.
Each case begins with five integers N, X, M, Y, K. (0 < N, M, X, Y, K < 109 and N ≠ M).
Output
For each case print one line containing the case number and L which is the desired length (mod 1000000007) or the string "Impossible" if it's not possible.
Sample Input
Output for Sample Input
2
3 16 5 42 6
5 1 6 10 9
Case 1: 68
Case 2: Impossible
PROBLEM SETTER: MD. TOWHIDUL ISLAM TALUKDER
SPECIAL THANKS: JANE ALAM JAN
题意：一个字符串，每次a变成b，b变成ab，第一个ab的数量不确定，现在给你第n次m次变换后的结果的字符串的长度，求第k次变换后字符串的长度（mod1e9+7）
题意没说清楚，第n和m次的长度应该都是没mod过1e9+7的，不然这题就复杂了
假设第一个里面a的数量为x，b的数量为y，变换一次后变为y，x+y，总长度为x+2y，变换两次为x+y，x+2y，总长度为2x+3y，第三次变换后为x+2y,2x+3y,总长度为3x+5y，可以发现x和y的系数是按斐波拉契数列变化的，解方程判断是否能求出x，y对应的系数即可，x y也不能为负
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
#define pi acos(-1.0)
#define eps 1e-10
#define pf printf
#define sf scanf
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define e tree[rt]
#define _s second
#define _f first
#define all(x) (x).begin,(x).end
#define mem(i,a) memset(i,a,sizeof i)
#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
#define mi ((l+r)>>1)
#define sqr(x) ((x)*(x))
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
ll p[3][3],ans[3][3];
int m,n,x,y,k,t,XX,Xx,xX,YY,Yy,yY;//定义成x1，y1什么的CE了，很迷
void multi(ll a[][3],ll b[][3])
{
ll tmp[3][3];
mem(tmp,0);
for1(i,2)
for1(j,2)
for1(k,2)
tmp[i][j]=(tmp[i][j]+a[i][k]*b[k][j])%mod;
for1(i,2)
for1(j,2)
a[i][j]=tmp[i][j];
}
void init()
{
mem(ans,0);
mem(p,0);
ans[1][1]=ans[2][2]=1;
p[1][1]=p[1][2]=p[2][1]=1,p[2][2]=0;
}
void solve(int n,int& x,int& y)//矩阵快速幂计算x y的系数
{
n--;
init();
while(n)
{
if(n&1)multi(ans,p);
n>>=1;
multi(p,p);
}
x=(ans[2][1]+ans[2][2])%mod;
y=(ans[1][1]+ans[1][2])%mod;
}
int main()
{
sf("%d",&t);
for1(i,t)
{
init();
int tag=0;
sf("%d%d%d%d%d",&x,&n,&y,&m,&k);
pf("Case %d: ",i);
solve(x,XX,YY);
solve(y,Xx,Yy);
solve(k,xX,yY);
if((XX*m-Xx*n)%(Yy*XX-YY*Xx))//不能整除，说明不能求出满足的x y，因为 m n没有mod过1e9+7，在x y大于0时要求出解，否则不满足
puts("Impossible");
else
{
ll YYY=(XX*m-Xx*n)/(Yy*XX-YY*Xx);
if((n-YY*YYY)%XX)
puts("Impossible");
else
{
int XXX=(n-YY*YYY)/XX;
if(XXX<0||YYY<0)
{
puts("Impossible");
}
else
{
ll ans=((ll)XXX*xX%mod+(ll)YYY*yY%mod)%mod;
ans=(ans+mod)%mod;
pf("%lld\n",ans);
}
}
}
}
return 0;
}
