| PDF (English) | Statistics | Forum |
| 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
思路：矩阵快速幂；
感觉这道题的题意有点问题，所给你的条件不知道是否取模，不过最后错了好几次，，和样例可以确定是没取模。
那么说下思路：Fa（n+1）=Fb(n);Fb (n+1)=Fa(n)+Fb(n);
然后给你两个条件，那么你可以设fa（1）=x；fb（1）=y；然后解方程组，然后判定下方程是否有解，其中x>=0&&y>=0;其中系数用矩阵快速幂算下。
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<string.h>
5 #include<stdlib.h>
6 #include<queue>
7 #include<math.h>
8 #include<vector>
9 using namespace std;
10 typedef long long LL;
11 const LL mod=1e9+7;
12 LL quick1(LL n,LL m)
13 {
14 LL a=1;
15 while(m)
16 {
17 if(m&1)
18 {
19 a=(a*n)%mod;
20 }
21 n=(n*n)%mod;
22 m/=2;
23 }
24 return a;
25 }
26 typedef struct pp
27 {
28 LL m[3][3];
29 pp()
30 {
31 memset(m,0,sizeof(m));
32 }
33 } maxtr;
34 void Init(maxtr *ans)
35 {
36 int i,j,k;
37 for(i=0; i<=1; i++)
38 {
39 for(j=0; j<=1; j++)
40 {
41 if(i==0&&j==0)
42 {
43 ans->m[i][j]=0;
44 }
45 else if(i==1&&j==1)
46 {
47 ans->m[i][j]=1;
48 }
49 else ans->m[i][j]=1;
50 }
51 }
52 }
53 maxtr E()
54 {
55 maxtr ans;
56 int i,j;
57 for(i=0; i<=1; i++)
58 {
59 for(j=0; j<=1; j++)
60 {
61 if(i==j)
62 {
63 ans.m[i][j]=1;
64 }
65 else ans.m[i][j]=0;
66 }
67 }
68 return ans;
69 }
70 maxtr quick( maxtr ans,LL m)
71 {
72 maxtr cc=E();
73 while(m)
74 {
75 if(m&1)
76 {
77 maxtr ak;
78 int s;
79 int i,j;
80 for(i=0; i<=1; i++)
81 {
82 for(j=0; j<=1; j++)
83 {
84 for(s=0; s<=1; s++)
85 {
86 ak.m[i][j]=(ak.m[i][j]+(cc.m[s][j]*ans.m[i][s])%mod)%mod;
87 }
88 }
89 }
90 cc=ak;
91 }
92 maxtr ak;
93 int s;
94 int i,j;
95 for(i=0; i<=1; i++)
96 {
97 for(j=0; j<=1; j++)
98 {
99 for(s=0; s<=1; s++)
100 {
101 ak.m[i][j]=(ak.m[i][j]+(ans.m[i][s]*ans.m[s][j])%mod)%mod;
102 }
103 }
104 }
105 ans=ak;
106 m/=2;
107 }
108 return cc;
109 }
110 LL gcd(LL n, LL m)
111 {
112 if(m==0)
113 {
114 return n;
115 }
116 else if(n%m==0)
117 {
118 return m;
119 }
120 else return gcd(m,n%m);
121 }
122 int main(void)
123 {
124 LL i,j,k;
125 LL s;
126 LL N, X, M, Y, K;
127 scanf("%d",&k);
128 LL acy;
129 LL acx;
130 for(s=1; s<=k; s++)
131 {
132 int flag=0;
133 scanf("%lld %lld %lld %lld %lld",&N,&X,&M,&Y,&K);
134 if(N>M)
135 {
136 swap(N,M);
137 swap(X,Y);
138 }
139 maxtr ak;
140 Init(&ak);
141 maxtr ac=quick(ak,N-1);
142 maxtr bk;
143 Init(&bk);
144 maxtr aak=quick(bk,M-1);
145 LL xx1=(ac.m[0][0]+ac.m[1][0])%mod;
146 LL yy1=(ac.m[0][1]+ac.m[1][1])%mod;
147 LL xx2=(aak.m[0][0]+aak.m[1][0])%mod;
148 LL yy2=(aak.m[0][1]+aak.m[1][1])%mod;
149 //printf("%lld %lld\n",xx1,xx2);
150 //printf("%lld %lld\n",yy1,yy2);
151 LL ccy=yy1;
152 LL xxN=X;
153 LL t=X;
154 LL xxy=yy2;
155 LL ct=Y;
156 LL yyM=Y;
157 LL gc=gcd(xx1,xx2);
158 yy1*=(xx2/gc);
159 yy2*=(xx1/gc);
160 X*=(xx2/gc);
161 Y*=(xx1/gc);
162 yy1-=yy2;
163 X-=Y;
164 if(X<0&&yy1>0||X>0&&yy1<0)flag=1;
165 {
166 if(X==0)
167 {
168 acy=0;
169 if(xxN%xx1)
170 {
171 flag=1;
172 }
173 else
174 {
175 LL ack=quick1(xx1,mod-2);
176 acx=xxN*ack%mod;
177 }
178 }
179 else
180 {
181
182 if(X%yy1)
183 {
184 flag=1;
185 }
186 else
187 { if(yy1>0&&X<0||yy1<0&&X>0)flag=1;
188 LL ack=quick1(yy1,mod-2);
189 acy=X/yy1%mod;
190 xxN-=ccy*acy;
191 if(xxN<0)flag=1;
192 if(xxN%xx1)
193 {
194 flag=1;
195 }
196 else
197 {
198 LL ack=quick1(xx1,mod-2);
199 acx=xxN/xx1;
200 }
201 }
202 }
203
204 }
205 printf("Case %d: ",s);
206 if(flag)
207 {
208 printf("Impossible\n");
209 }
210 else
211 {
212 maxtr cp;
213 Init(&cp);
214 maxtr akk=quick(cp,K-1);
215 LL xxx1=(akk.m[0][0]+akk.m[1][0])*acx%mod;
216 LL yyy1=(akk.m[0][1]+akk.m[1][1])*acy%mod;
217 printf("%lld\n",(xxx1+yyy1)%mod);
218 }
219 }
220 return 0;
221 }